Prove Mathematics

Cauchy Sequence

noreply@blogger.com (Sampark Sathi) — Tue, 27 Jan 2015 20:29:00 +0000

Definition: Cauchy sequence is a sequence whose elements become arbitrarily close to each other as the sequence progresses. It can be defined for various spaces, such as:

For real numbers, a sequence $\langle x_n\rangle$ of real numbers, such that:

$\qquad\quad\langle x_n\rangle = x_1, x_2, x_3, \cdots$ where, $x_1, x_2, x_3, \cdots\;\in\;\mathbb{R}$

is a Cauchy sequence, if for every positive real number $\epsilon$, $\exists$ an $N\;\in\;\mathbb{N}$, such that

$\qquad\quad |x_n - x_m| < \epsilon \;\;\forall\;\; n, m\;\geq\; N$

Similarly, for rational numbers, a sequence $\langle x_n\rangle$ of rational numbers, such that:

$\qquad\quad\langle x_n\rangle = x_1, x_2, x_3, \cdots$ where, $x_1, x_2, x_3, \cdots\;\in\;\mathbb{Q}$

is a Cauchy sequence, if for every positive rational number $\epsilon$, $\exists$ an $N\;\in\;\mathbb{N}$, such that

$\qquad\quad |x_n - x_m| < \epsilon \;\;\forall\;\; n, m\;\geq\; N$

For example:

$\textrm{If}\;\;\;\quad \langle x_n\rangle =\left\{\dfrac{1}{n}\;\big|\;\; n\;\in\;\mathbb{Z}\right\}\\
\begin{align}\textrm{Then,}\; |x_n - x_m| &= \left| \dfrac{1}{n} - \dfrac{1}{m}\right|\\[6pt]
& < \dfrac{1}{n} \text{&} \dfrac{1}{m}\end{align}$

Choosing:
$\begin{align}\qquad\quad & N > \dfrac{1}{\epsilon}\\
\Rightarrow\quad\;\; &\dfrac{1}{N} < \epsilon\\
\therefore\quad\;\;\; &|x_n - x_m|\;\;< \; \dfrac{1}{n} \text{&} \dfrac{1}{m} < \;\epsilon\;\;\forall\;\; n, m\;\geq\; N \;\left(\textrm{for}\; N > \dfrac{1}{\epsilon}\right) \end{align}$

Hence, $\langle x_n\rangle =\left\{\dfrac{1}{n}\;\big|\;\; n\;\in\;\mathbb{Z}\right\}$ is a Cauchy sequence.

Abelian Group

noreply@blogger.com (Sampark Sathi) — Tue, 27 Jan 2015 20:18:00 +0000

Definition: An Abelian group, or a commutative group, is a group which along with the group axioms, satisfies an additional axiom of commutativity, i.e., if $(G, •)$ is an abelian group, then it satisfies the following five axioms:

Closure:
$\forall\;\;a, b\;\in\; G, \; a • b \;\in\; G$.

Associativity:
$\forall\;\;a, b$ and $c\;\in\; G, \; (a • b) • c = a • (b • c)$.

Existence of identity:
$\exists\;$ an element $e\;\in\; G$, such that $\forall\; a\;\in\; G,$

$\qquad\quad e • a = a • e = a$

Such $e$ is known as the identity element of $G$ w.r.t. $•$.

Existence of inverse:
For each $a\;\in\; G$, $\exists$ an element $b\;\in\; G$, such that,

$\qquad\quad a • b = b • a = e$

Where $e$ is the identity element, and $b$ is called the inverse of $a$ in $(G, •)$.

Commutativity:
$\forall\;\;a, b\;\in\; G$,

$\qquad\quad a • b = b • a$

Group

noreply@blogger.com (Sampark Sathi) — Tue, 27 Jan 2015 20:05:00 +0000

Definition: A group is a set, $G$, together with an operation $•$ that combines any two elements $a$ and b to form another element, denoted by $a • b$ or $ab$. A group $(G, •)$, must satisfy four requirements known as the group axioms:

Closure:
$\forall\;\;a, b\;\in\; G, \; a • b \;\in\; G$.

Associativity:
$\forall\;\;a, b$ and $c\;\in\; G, \; (a • b) • c = a • (b • c)$.

Existence of identity:
$\exists\;$ an element $e\;\in\; G$, such that $\forall\; a\;\in\; G,$

$\qquad\quad e • a = a • e = a$

Such $e$ is known as the identity element of $G$ w.r.t. $•$.

Existence of inverse:
For each $a\;\in\; G$, $\exists$ an element $b\;\in\; G$, such that,

$\qquad\quad a • b = b • a = e$

Where $e$ is the identity element, and $b$ is called the inverse of $a$ in $(G, •)$.

Rational Numbers

noreply@blogger.com (Sampark Sathi) — Tue, 27 Jan 2015 19:19:00 +0000

Definition: a rational number is any number that can be written in the p/q form of two integers, p and q, with the denominator q not equal to zero. The set of all rational numbers is denoted by $\mathbb{Q}$, i.e.,

$\qquad\quad\mathbb{Q} = \left\{ \dfrac{p}{q} \;\big|\;\; p, q\;\in\;\mathbb{Z};\;\; q\neq 0\right\}$

In decimal representation, a number is a rational number if and only if the decimal expansion of a number either terminates after a finite number of digits or begins to repeat the same finite sequence of digits over and over. This holds true irrespective of the base, i.e, decimal, binary, etc.

Related:
Integers (definition)

Integers

noreply@blogger.com (Sampark Sathi) — Tue, 27 Jan 2015 19:08:00 +0000

Definition: An integer is a number that can be written without a fractionl component. The set of integers is denoted by $\mathbb{Z}$. Further, if $n$ is an integer, then $n^+$ (or the successor of $n$, eg. $0^+ = 1$, $1^+ = 2$) is also an integer. Similarly, if $n$ is an integer, then $n^-$ is also an integer.

Median Through Hypotenuse

noreply@blogger.com (Sampark Sathi) — Wed, 14 Jan 2015 23:05:00 +0000

Theorem: In a right triangle, length of a median drawn through the vertex having right angle to meet hypotenuse, is equal to one half of the length of the hypotenuse.

Prerequisites:
Median (definition)
Midpoint Theorem (proof)
SAS congruence (proof)
Angle on a straight line (proof)
Corresponding angles property (proof)

Proof:

Let $\triangle ABC$ be a right triangle, right angled at B. Let BD be a median drawn from B to meet AC at D.

We need to show that BD = $\dfrac{1}{2}$ AC. For this, let us join DE, where E is the midpoint of AB.

Since, D and E are the midpoints of AC and AB respectively (see definition of median), hence by Midpoint Theorem,

$\begin{align}\qquad\quad &DE \parallel BC\\
\Rightarrow\quad\;\; &\angle AED = \angle ABC = 90^o\quad\qquad\qquad\!\! && \text{(corresponding angles)} &&& \cdots\text{(1)}\\
\Rightarrow\quad\;\; & \angle BED = 180^o -\angle AED = 90^o && \text{(angle on a straight line)} &&& \cdots\text{(2)}\end{align}$

Now, consider $\triangle AED$ and $\triangle BED$:

$\begin{align}\qquad\quad\;\; & AE = BE\qquad\qquad\qquad\qquad\qquad && \text{(by construction)}\\
& DE = DE && \text{(common)}\\
& \angle AED = \angle BED = 90^o && \text{(from $(1)$ and $(2)$)}\\
\therefore\quad\;\;\;\;\; & \triangle AED\cong\triangle BED && \text{(by SAS congruence)}\\
\\[12pt]
\Rightarrow\;\quad\quad & AD = BD && \text{(CPCTC)}\qquad\qquad\qquad\qquad\qquad\cdots\text{(3)}\\
\text{But, }\;\quad & AD = AC/2 && \text{(by definition of median)}\\
\text{Hence, }\; & BD = AC/2 && \text{(from $(3)$)}\end{align}$

Q.E.D.

Recommended;
Location of centroid
Pythagoras Theorem
Angle Bisector Theorem

Centroid

noreply@blogger.com (Sampark Sathi) — Wed, 14 Jan 2015 22:50:00 +0000

Theorem: All the three medians of a triangle intersect at a single point called centroid, which divides each median in a ratio of $2:1$.

Prerequisites:
Similarity of triangles (definition)
Midpoint Theorem (proof)
AA similarity (proof)
Alternate angles property (proof)
vertical angle theorem (proof)

Proof:

Let ABC be a triangle having medians AD, BE and CF.
Let us consider the medians BE and CF, which intersect at a point G. Let us join E and F by a straight line.

Since EF joins the midpoints of the lines AB and AC, hence by midpoint theorem:

$\qquad\quad EF \parallel BC$
And, $\;\;\;\! EF = \dfrac{1}{2} BC\qquad\qquad\qquad\cdots\text{(1)}$

Now, consider triangles $\triangle BCG$ and $\triangle EFG$:

$\qquad\quad\angle EGF = \angle BGC\qquad\qquad\qquad\text{(vertically opposite angles)}\\
\qquad\quad\angle GFE = \angle GCB\qquad\qquad\qquad\text{(alternate angles)}\\
\therefore\quad\;\;\triangle BCG\sim\triangle EFG\qquad\qquad\qquad\text{(AA similarity)}$

Thus, by definition of similarity, corresponding sides are proportional, hence:

$\qquad\quad\dfrac{GE}{GB} = \dfrac{GF}{GC} =\dfrac{EF}{BC} = \dfrac{1}{2}\qquad\qquad\text{(from $(1)$)}$

Thus, G divides BE and CF in the ratio $2:1$.

Similarly, by considering the medians AD and BE, which intersect at a point G', it can be shown that G' divides AD and BE in the ratio 2:1.

But BE is divided in $2:1$ ratio by G. Hence, G' $=$ G.

Thus, all the medians intersect at a single point, which divides the medians in a ratio $2:1$.

Q.E.D.

Recommended;
Median through hypotenuse
Basic Proportionality Theorem
Angle Bisector Theorem

Median

noreply@blogger.com (Sampark Sathi) — Wed, 14 Jan 2015 22:46:00 +0000

Definition: In a triangle, median is a line drawn from a vertex to join the midpoint of the opposite side.

Converse Of Pythagoras Theorem

noreply@blogger.com (Sampark Sathi) — Wed, 14 Jan 2015 22:37:00 +0000

Theorem: in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.

Prerequisites:
Pythagoras theorem (proof)
SSS congruence (proof)

Proof:

Let there be a $\triangle ABC$, such that:

$\qquad\quad AB^2 + BC^2 = AC^2$

We need to prove that $\triangle ABC$ is a right triangle. For this, we construct a right triangle $\triangle PQR$, right angled at Q, such that PQ $=$ AB and QR $=$ BC.

Since, $\triangle PQR$ is a right triangle, hence by using pythagoras theorem, we get:

$\qquad\quad PQ^2 + QR^2 = PR^2\\
\Rightarrow\quad\;\; AB^2 + BC^2 = PR^2\qquad\qquad\qquad\text{(by construction)}\qquad\qquad\!\cdots\text{(1)}$
But, $\quad AB^2 + BC^2 = AC^2\qquad\qquad\qquad\text{(given)}\qquad\qquad\qquad\qquad\;\cdots\text{(2)}$

From $(1)$ and $(2)$,

$\qquad\quad PR^2 = AC^2\\
\Rightarrow\quad\;\; PR = AC$

Also, since PQ $=$ AB and QR $=$ BC by construction, hence by SSS congruency rule,

$\qquad\quad\triangle ABC\cong\triangle PQR$
$\Rightarrow\quad\;\;\angle B = \angle Q$
But, $\quad\angle Q = 90^o$
$\therefore\quad\;\;\;\angle B = 90^o$

Thus, $\triangle ABC$ is a right triangle, right angled at B.

Q.E.D.

Pythagoras Theorem

noreply@blogger.com (Sampark Sathi) — Wed, 14 Jan 2015 21:50:00 +0000

Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Method 1: By similarity:

Prerequisites:
Similarity of triangles (definition)
AA similarity (proof)

Proof:

Let $\triangle ABC$ be a right triangle, right angled at B. Let us draw an altitude from vertex B to the line AC, meeting AC at a point D.

Now, in $\triangle ADB$ and $\triangle ABC$,

$\qquad\quad\:\!\angle A = \angle A\qquad\qquad\qquad\qquad\quad\:\text{(common)}\\
\qquad\quad\angle ADB = \angle ABC = 90^o\\
\therefore\quad\;\;\;\triangle ADB\sim\triangle ABC\qquad\qquad\qquad\text{(by AA similarity rule)}$

Since, by definition of similar triangles, corresponding sides of similar triangles are proportional,

$\therefore\quad\;\;\dfrac{AD}{AB} = \dfrac{AB}{AC}\\[12pt]
\Rightarrow\quad\;\; AB^2 = AD \times AC\qquad\qquad\qquad\cdots\text{(1)}$

Similarly, from $\triangle BDC$ and $\triangle ABC$,

$\qquad\quad BC^2 = DC \times AC\qquad\qquad\qquad\cdots\text{(2)}$

Adding $(1)$ and $(2)$,

$\qquad\quad\begin{align} AB^2 + BC^2 &= AD\times AC + DC\times AC\\
&= (AD + DC)\times AC\\
&= AC\times AC\\
&= AC^2\end{align}$

Q.E.D.

Method 2: By Area:

Prerequisites:
Square (definition)
Area of square (proof)
Area of right triangle (proof)
SAS congruence (proof)
Angle sum property of triangle (proof)
Angle on a straight line (proof)
$(a+b)^2 = a^2 + b^2$ (proof)

Proof:

Let ABCD be a square having side length equal to $(a+b)$. Let us locate the points E, F, G and H on the lines AB, BC, CD and DA respectively, such that AE $=$ BF $=$ CG $=$ DH $=$ $a$. Join EF, FG, GH and HE by straight lines, as shown in the figure.

Let $\angle AEH = \theta$, then by angle sum property of a triangle,

$\qquad\quad\begin{align}\angle AHE &= 180^o - \angle A - \angle AEH\\
&= 180^o - 90^o - \theta\\
&= 90 -\theta\qquad\qquad\qquad\qquad\cdots\text{(1)}\end{align}$

Now, in $\triangle HAE$ and $\triangle EBF$,

$\qquad\quad AE = BF = a\qquad\qquad\qquad\text{(by construction)}\\
\qquad\quad AH = BE = b\qquad\qquad\qquad\text{(by construction)}\\
\qquad\quad\angle A = \angle B = 90^o\qquad\qquad\quad\:\!\text{(by definition of square)}$

Hence, $\triangle HAE\cong\triangle EBF$ by SAS rule.

Thus, by CPCTC,

$\qquad\quad\angle BEF = \angle AHE = 90^o - \theta\qquad\qquad\qquad\text{(from $(1)$)}\\
\begin{align}\therefore\quad\;\;\;\angle HEF &= 180^o - \angle AEH - \angle BEF\qquad\quad\text{($\because$ AB is a straight line)}\\
&= 180^o - \theta - (90^o - \theta)\\
&= 90^o\qquad\qquad\qquad\qquad\qquad\qquad\;\cdots\text{(2)}\end{align}$

Similarly, all the four triangles, i.e., $\triangle HAE,\;\triangle EBF,\;\triangle FCG$ and $\triangle GDH$ are congruent. Thus, by CPCTC, $\angle HEF, \; \angle EFG,\; \angle FGH$ and $\angle GHE$ are equal and, from $(2)$, are equal to $90^o$. Also, EF $=$ FG $=$ GH $=$ HE $=$ $c$ (say). Hence, EFGH is a square (see definition of a square).

Further, since these four triangles are congruent, they have same area.

Now, let us find the area of the square ABCD:

$\qquad\quad \text{Area of }\; ABCD = \text{Area of } \;EFGH + ar (\triangle HAE) + ar (\triangle EBF)+ ar (\triangle FCG) \\
\\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad + ar (\triangle GDH)\\
\Rightarrow\quad\;\; (a+b)^2 = c^2 + 4\times\frac{1}{2} ab\qquad\qquad\;\text{(by formula for area of square and right triangle)}\\
\Rightarrow\quad\;\; a^2 + b^2 + 2ab = c^2 + 2ab\qquad\qquad\text{($\because\;\;(a+b)^2 = a^2 + b^2 + 2ab$)}\\
\Rightarrow\quad\;\; a^2 + b^2 = c^2\qquad\qquad\qquad\qquad\quad\;\;\cdots\text{(3)}$

Since, $a$ and $b$ are chosen arbitrarily, hence the result of equation $(3)$ holds for all right triangles.

Q.E.D.

SAS Similarity

noreply@blogger.com (Sampark Sathi) — Wed, 14 Jan 2015 21:23:00 +0000

Theorem: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

Prerequisites:
AA similarity (proof)
Converse of Basic Proportionality Theorem (proof)
SAS congruence (proof)
Corresponding angles property (proof)

Proof:

Let there be two triangles $\triangle ABC$ and $\triangle DEF$, such that:

$\qquad\quad\dfrac{AB}{DE} = \dfrac{AC}{DF}\\[12pt]
\text{And, }\;\;\angle A = \angle D$

In order to prove that $\triangle ABC\sim\triangle DEF$, let us draw a line PQ, where P lies on the line DE and Q lies on the line DF, such that DP $=$ AB and DQ $=$ AC.

Now in triangles $\triangle ABC$ and $\triangle DPQ$,

$\qquad\quad AB = DP\qquad\qquad\qquad\text{(by construction)}\\
\qquad\quad AC = DQ\qquad\qquad\qquad\text{(by construction)}\\
\qquad\quad\angle A = \angle D\qquad\qquad\qquad\:\!\!\:\text{(given)}$

Hence, $\triangle ABC\cong\triangle DPQ$ by SAS rule of congruence.

Now,
$\qquad\quad\!\dfrac{AB}{DE} = \dfrac{AC}{DF}\qquad\qquad\qquad\:\!\text{(given)}\\[12pt]
\therefore\;\;\;\quad\dfrac{DP}{DE} = \dfrac{DQ}{DF}\qquad\qquad\qquad\text{(by construction)}\\[12pt]
\therefore\;\;\;\quad PQ \parallel EF\qquad\qquad\qquad\quad\;\;\!\text{(by converse of Basic Proportionality Theorem)}\\
\therefore\;\;\;\quad\angle DPQ = \angle E\qquad\qquad\qquad\!\!\text{(corresponding angles)}$

Also, since $\angle D$ is common in $\triangle DPQ$ and $\triangle DEF$, hence by AA similarity,

$\qquad\quad\!\triangle DPQ\sim\triangle DEF\\
\therefore\quad\;\;\;\triangle ABC\sim\triangle DEF\qquad\qquad\qquad\text{(as $\triangle ABC\cong\triangle DPQ$)}$

Q.E.D.

Recommended:
SSS similarity
Midpoint Theorem
Pythagoras Theorem

SSS Similarity

noreply@blogger.com (Sampark Sathi) — Wed, 14 Jan 2015 21:08:00 +0000

Theorem: If the corresponding sides of two triangles are proportional, then they are similar.

Prerequisites:
Similarity of triangles (definition)
AA Similarity (proof)
Converse of Basic Proportionality Theorem (proof)
SSS congruence (proof)
Corresponding angles property (proof)

Proof:

Let there be two triangles $\triangle ABC$ and $\triangle DEF$, such that:

$\qquad\quad\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$

In order to prove that $\triangle ABC\sim\triangle DEF$, let us draw a line PQ, where P lies on the line DE and Q lies on the line DF, such that DP $=$ AB and DQ $=$ AC.

$\text{Since, }\;\dfrac{AB}{DE} = \dfrac{AC}{DF}\\[12pt]
\text{Hence, }\dfrac{DP}{DE} = \dfrac{DQ}{DF}\qquad\qquad\qquad\text{(by construction)}\\[12pt]
\therefore\quad\;\;\;\;\: PQ \parallel EF\qquad\qquad\qquad\;\quad\text{(by converse of Basic Proportionality Theorem)}\\
\Rightarrow\quad\;\;\:\angle DPQ = \angle E\qquad\qquad\qquad\!\!\:\!\text{(corresponding angles)}$

Also, since $\angle D$ is common in $\triangle DPQ$ and $\triangle DEF$, thus, $\triangle DPQ\sim\triangle DEF\;$ by AA similarity.

Therefore, by the definition of similarity of triangles,

$\qquad\quad\dfrac{DP}{DE} = \dfrac{PQ}{EF}\\[12pt]
\Rightarrow\quad\;\;\dfrac{AB}{DE} = \dfrac{PQ}{EF}\qquad\qquad\qquad\text{(by construction)}\qquad\cdots\text{(1)}\\[12pt]
\text{But, }\;\;\dfrac{AB}{DE} = \dfrac{BC}{EF}\qquad\qquad\qquad\text{(given)}\qquad\qquad\qquad\;\:\cdots\text{(2)}\\[12pt]
\Rightarrow\quad\;\;\:\!\dfrac{PQ}{EF}=\dfrac{BC}{EF}\qquad\qquad\qquad\!\!\:\text{(from $(1)$ and $(2)$)}\\[12pt]
\Rightarrow\quad\;\;\; PQ = BC\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\cdots\text{(3)}$

Now, in $\triangle ABC$ and $\triangle DPQ$,

$\qquad\quad AB = DP\qquad\qquad\qquad\text{(by construction)}\\
\qquad\quad AC = DQ\qquad\qquad\qquad\:\!\!\text{(by construction)}\\
\qquad\quad BC = PQ\qquad\qquad\qquad\text{(from $(3)$)}$

Hence, $\triangle ABC\cong\triangle DPQ$ by SSS rule.

Thus, since

$\qquad\quad\!\triangle DPQ\sim\triangle DEF\\
\therefore\quad\;\;\;\triangle ABC\sim\triangle DEF$

Q.E.D.

Recommended:
SAS similarity
Midpoint Theorem
Pythagoras Theorem

AA Similarity

noreply@blogger.com (Sampark Sathi) — Tue, 13 Jan 2015 18:20:00 +0000

Theorem: Two triangles are similar if any two angles of one triangle are respectively equal to two angles of the other triangle.

Prerequisites:
AAA similarity (proof)
Angle sum property of triangle (proof)

Proof:

Let $\triangle ABC$ and $\triangle DEF$ be two triangles such that $\angle A = \angle D$ and $\angle B = \angle E$.
By using angle sum property of triangles,

$\qquad\quad\begin{align}\angle C &= 180^o - \angle A - \angle B\\
&= 180^o - \angle D -\angle E\qquad\qquad\text{(given)}\\
&= \angle F\end{align}$

Hence, all the three pair of angles are congruent. Thus, $\triangle ABC\sim\triangle DEF$ by AAA similarity.

Recommended:
SSS similarity
Midpoint Theorem
Angle Bisector Theorem

AAA Similarity

noreply@blogger.com (Sampark Sathi) — Tue, 13 Jan 2015 18:12:00 +0000

Theorem: If in two triangles, corresponding angles are equal, then the triangles are similar.

Prerequisites:
Similarity of triangles (definition)
Basic Proportionality Theorem (proof)
Alternate angle property (proof)
SAS congruence (proof)

Proof:

Let there be two triangles $\triangle ABC$ and $\triangle DEF$, such that $\angle A = \angle D$, $\angle B = \angle E$ and $\angle C = \angle F$.

We need to show that $\triangle ABC\sim\triangle DEF$. For this draw a line PQ, where P and Q lies on the lines DE and DF respectively, such that AB $=$ DP and AC $=$ DQ.

In $\triangle ABC$ and $\triangle DPQ$,

$\qquad\quad AB = DP\qquad\qquad\qquad\qquad\text{(by construction)}\\
\qquad\quad AC = DQ\qquad\qquad\qquad\qquad\text{(by construction)}\\
\qquad\quad\angle A = \angle D\qquad\qquad\qquad\qquad\:\!\text{(given)}$

Hence, $\triangle ABC\cong\triangle DPQ$ by SAS rule.

Therefore, by CPCTC,

$\;\qquad\quad\angle B = \angle DPQ$
But, $\;\quad\angle B = \angle E\qquad\qquad\qquad\qquad\text{(given)}$
Hence, $\;\angle DPQ = \angle E$

Thus, by alternate angle property, PQ $\parallel$ EF.

Now using Basic Proportionality theorem in $\triangle DEF$,

$\qquad\quad\dfrac{PE}{DP} = \dfrac{QF}{DQ}\\[12pt]
\Rightarrow\quad\;\;\dfrac{PE}{DP} + 1 = \dfrac{QF}{DQ} + 1\\[12pt]
\Rightarrow\quad\;\;\dfrac{DP + PE}{DP} = \dfrac{DQ + QF}{DQ} \\[12pt]
\Rightarrow\quad\;\;\dfrac{DE}{DP}= \dfrac{DF}{DQ}\\[12pt]
\Rightarrow\quad\;\;\dfrac{DE}{AB} = \dfrac{DF}{AC}\qquad\qquad\qquad\qquad\text{(by construction)}$

Similarly,

$\qquad\quad\dfrac{DE}{AB}= \dfrac{EF}{BC}$

Hence,

$\qquad\quad\dfrac{DE}{AB}=\dfrac{DF}{AC}=\dfrac{EF}{BC}$

Thus, $\triangle ABC\sim\triangle DEF$ by the definition of similarity of triangles.

Recommended:
SSS similarity
Midpoint Theorem
Angle Bisector Theorem

Angle Bisector Theorem

noreply@blogger.com (Sampark Sathi) — Tue, 13 Jan 2015 14:31:00 +0000

Theorem: Bisector of an angle of a trinagle divides the opposite sides in the ratio of the sides containing the angle.

Prerequisites:
Basic Proportionality Theorem (proof)
Alternate angles property (proof)
Corresponding angles property (proof)
Converse of Isosceles Triangle Theorem (proof)

Proof:

Let ABC be a triangle with AD being the bisector of $\angle A$, meeting BC at D. We need to show that:

$\qquad\quad\dfrac{BD}{DC} = \dfrac{AB}{AC}$

For this, let us draw CE $\parallel$ DA to meet the extended BA at C. Since CE $\parallel$ DA,

$\therefore\quad\;\;\;\angle CAD = \angle ACE\qquad\qquad\quad\text{(Alternate angles)}\qquad\qquad\qquad\qquad\qquad\;\;\:\cdots\text{$(1)$}\\[6pt]
\text{Also, }\;\:\!\angle BAD = \angle AEC\qquad\quad\qquad\text{(Corresponding angles)}\qquad\qquad\qquad\qquad\;\;\cdots\text{$(2)$}\\[6pt]
\text{But, }\;\;\angle BAD = \angle CAD\quad\qquad\qquad\:\!\:\!\!\text{(As AD bisects $\angle A$)}\\[6pt]
\therefore\quad\;\;\;\angle ACE = \angle AEC\quad\qquad\!\:\qquad\text{(from $(1)$ and $(2)$)}\\[6pt]
\Rightarrow\quad\;\; AC = AE\qquad\qquad\qquad\qquad\text{(Converse of Isosceles Triangle Theorem)}\quad\cdots\text{$(3)$}$

Now, in $\triangle BCE$, DA $\parallel$ CE. Thus, by Basic Proportionality Theorem,

$\qquad\quad\dfrac{BD}{DC} = \dfrac{BA}{AE}\\[8pt]
\Rightarrow\quad\;\;\dfrac{BD}{DC} = \dfrac{AB}{AC}\qquad\qquad\qquad\;\;\text{(from $(3)$)}$

Q.E.D.

Recommended:
Basic Proportionality Theorem
Midpoint Theorem
Location of centroid

Similarity Of Triangles

noreply@blogger.com (Sampark Sathi) — Tue, 13 Jan 2015 14:25:00 +0000

Definition: Two triangles are said to be similar if and only if their corresponding angles are equal, and their corresponding sides are proportional.

Converse Of Basic Proportionality Theorem

noreply@blogger.com (Sampark Sathi) — Tue, 13 Jan 2015 14:12:00 +0000

Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Prerequisites:
Basic Proportionality Theorem (proof)
Unique parallel through a point (proof)

proof:

Let there be a $\triangle ABC$ and a line $l$ intersecting the sides AB and AC at the points D and E respectively, as shown in the figure, such that:

$\qquad\quad\dfrac{AD}{DB} = \dfrac{AE}{EC}$

We need to show that $l \parallel BC$. For this, let us assume that $l \not\parallel BC$. Then there must exist a unique line DF through the point D, such that DF $\parallel$ BC.

Since DF $\parallel$ BC, hence by Basic Proportionality Theorem, we get:

$\qquad\quad\;\:\!\dfrac{AD}{DB} = \dfrac{AF}{FC}\\[12pt]
\text{But, }\quad\dfrac{AD}{DB} = \dfrac{AE}{EC}\qquad\qquad\qquad\qquad\text{(given)}\\[12pt]
\therefore\quad\quad\:\dfrac{AF}{FC} = \dfrac{AE}{EC}\\[12pt]
\Rightarrow\quad\;\;\;\dfrac{AF}{FC} +1 = \dfrac{AE}{EC} +1\\[12pt]
\Rightarrow\quad\;\;\;\dfrac{AF+FC}{FC} = \dfrac{AE+EC}{EC}\\[12pt]
\Rightarrow\quad\;\;\;\:\dfrac{AC}{FC} = \dfrac{AC}{EC}\\[12pt]
\Rightarrow\qquad\; FC = EC$

But this is a contradiction as F $\neq$ E. Hence, our assumption was false. Therefore, $l \parallel BC$.

Q.E.D.

Recommended:
Angle Bisector Theorem
Midpoint Theorem
Location of centroid

Basic Proportionality Theorem

noreply@blogger.com (Sampark Sathi) — Tue, 13 Jan 2015 13:49:00 +0000

Theorem: In a triangle, a line drawn parallel to one side, to intersect the other sides in distinct points, divides the two sides in the same ratio.

Prerequisites:
Area of triangle (proof)
Equidistance property of parallel lines (proof)

Proof:

Let there be a triangle $\triangle ABC$. Let DE be a line such that DE $\parallel$ BC, where points D and E lie on the sides AB and AC respectively.

In order to prove the theorem, let us join BE and CD by straight lines. Draw EF $\bot$ AB and DG $\bot$ AC.

Now consider the ratio:

$\qquad\quad\dfrac{ar (\triangle ADE)}{ar (\triangle BDE)}$

Where, $ar (\triangle ADE)$ implies the area of $\triangle ADE$.

From the figure, using formula for area of triangle, taking EF as a perpendicular on AB,

$\qquad\quad\dfrac{ar (\triangle ADE)}{ar (\triangle BDE)} = \dfrac{\frac{1}{2}(AD).(EF)}{\frac{1}{2}(DB).(EF)} = \dfrac{AD}{DB}\qquad\qquad\qquad\qquad\cdots\text{(1)}$

Also, by taking DG as a perpendicular on the side AC,

$\qquad\quad\dfrac{ar (\triangle ADE)}{ar (\triangle CDE)} = \dfrac{\frac{1}{2}(AE).(DG)}{\frac{1}{2}(EC).(DG)} = \dfrac{AE}{EC}\qquad\qquad\qquad\qquad\cdots\text{(2)}$

Again, draw perpendiculars from points D and E on the line BC at the points M and N respectively. Since DE $\parallel$ BC, hence, by equidistance property of parallel lines, DM $=$ EN.

$\begin{align}\therefore\quad\;\; ar (\triangle BDC) &= \dfrac{1}{2}(DM)(BC)\\
&= \dfrac{1}{2}(EN)(BC)\\
&= ar (\triangle BEC)\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;\cdots\text{(3)}\end{align}$

Now, consider area of $\triangle BDE$,

$\qquad\quad\begin{align} ar (\triangle BDE) &= ar (BDEC) - ar (\triangle BEC)\\
&= ar (BDEC) - ar (\triangle BDC)\qquad\qquad\qquad\quad\text{(from $(3)$)}\\
&= ar (\triangle CDE)\qquad\qquad\qquad\qquad\qquad\qquad\quad\cdots\text{(4)}\end{align}$

Using the result of $(4)$,

$\qquad\quad\dfrac{ar (\triangle ADE)}{ar (\triangle BDE)} = \dfrac{ar (\triangle ADE)}{ar (\triangle CDE)}$

Hence from $(1)$ and $(2)$

$\qquad\quad\dfrac{AD}{DB} = \dfrac{AE}{EC}$

Hence the result.

Converse Of Isosceles Triangle Theorem

noreply@blogger.com (Sampark Sathi) — Tue, 13 Jan 2015 13:39:00 +0000

Theorem: Sides opposite to the equal angles in a triangle are equal.

Prerequisites:
AAS congruency (proof)

Proof:

Let ABC be a triangle having $\angle B = \angle C$. Let us draw AD which bisects the $\angle A$ and meets BC at D.

In $\triangle ABD$ and $\triangle ACD$,

$\qquad\quad\angle B = \angle C\qquad\qquad\qquad\;\:\!\:\!\qquad\text{(given)}\\
\qquad\quad\angle BAD = \angle CAD\qquad\qquad\quad\:\text{(by construction)}$
Also, $\quad AD = AD\qquad\qquad\qquad\qquad\text{(common)}$

Hence, $\triangle ABD\cong\triangle ACD$ by AAS rule.

Thus, by CPCTC, AB = AC, i.e, $\triangle ABC$ is an isosceles triangle.

Q.E.D.

Recommended:
Isosceles Triangle Theorem
Angle Bisector Theorem
Location of centroid

AAS Congruence

noreply@blogger.com (Sampark Sathi) — Tue, 13 Jan 2015 13:27:00 +0000

Theorem: If two pairs of angles of two triangles are equal in measurement, and a pair of corresponding non-included sides are equal in length, then the triangles are congruent.

Prerequisites:
ASA congruency (proof)
Angle sum property of triangle (proof)

Proof:

Let there be two triangles $\triangle ABC$ and $\triangle DEF$, such that the angles $\angle ABC = \angle DEF$, $\angle ACB = \angle DFE$ and the non-included sides AB $=$ DE.

Using angle sum property of a triangle,

$\qquad\quad\begin{align}\angle BAC &= 180^o - \angle ABC - \angle ACB\\
&= 180^o - \angle DEF - \angle DFE\qquad\;\;\!\qquad\text{(given)}\\
&= \angle EDF\qquad\qquad\qquad\qquad\qquad\qquad\text{(by angle sum property)}\end{align}$

Hence, $\triangle ABC\cong\triangle DEF$ by ASA rule.

Thus, the given triangles are congruent.

Recommended:
RHS congruence
SSS congruence
AAA similarity

Equidistance Property Of Parallel Lines

noreply@blogger.com (Sampark Sathi) — Tue, 13 Jan 2015 13:06:00 +0000

Theorem: Perpendicular distance between the two parallel lines is constant.

Prerequisites:
Interior angle property of parallel lines (proof)
Alternate angle property (proof)
AAS congruency (proof)

Proof:

Let $l_1$ and $l_2$ be two parallel lines. Let AC and BD be two perpendicular drawn from $l_1$ to $l_2$.

Since, for parallel lines, any two consecutive interior angles are supplementary, hence all the four angles in the figure, i.e., $\angle A,\; \angle B, \;\angle C\; \text{and}\; \angle D$ are $90^o$.

Now, consider $\triangle ABC$ and $\triangle DCB$:
$\qquad\quad\angle A = \angle D\qquad\qquad\qquad\qquad\qquad\:\!\text{(Right angles)}\\
\qquad\quad\angle ABC = \angle BCD\qquad\quad\qquad\qquad\text{(Alternate angles)}\\
\qquad\quad BC = BC\qquad\qquad\qquad\qquad\qquad\text{(Common)}$

Hence, $\triangle ABC\cong\triangle DCB$ by AAS rule.

Thus, by CPCTC, AC = BD.

Q.E.D.

Recommended:
Unique parallel through a point
Corresponding angles property
Unique RHS triangle

Area of Square

noreply@blogger.com (Sampark Sathi) — Tue, 13 Jan 2015 13:01:00 +0000

Theorem: Area of a square is equal to the square of its side length.

Prerequisites:
Square (definition)
Area of rectangle (proof)

Proof:

Let ABCD be a square having side length equal to '$s$'. Since by definition, square is a special type of rectangle, hence using the formula for area of a rectangle, area of ABCD is:

$\qquad\quad\begin{align}\text{Area of ABCD} &= \text{(length)}\times\text{(breadth)}\\
&= s\times s\\
&= s^2\end{align}$

Q.E.D.

Recommended:
Area of triangle
Area of rectangle
characteristics of parallelogram

Area Of Rhombus

noreply@blogger.com (Sampark Sathi) — Tue, 13 Jan 2015 12:43:00 +0000

Theorem: Area of a rhombus is equal to one half of the product of diagonals.

Prerequisites:
Properties of rhombus (proof)
Area of right triangle (proof)

Proof:

Let ABCD be a rhombus having diagonals AC and BD of lengths $d_1$ and $d_2$ respectively, intersecting at a point E.
Since in a rhombus, diagonals bisect each other at right angles. Hence,

$\qquad\quad\; AE = CE = \dfrac{d_1}{2}\\
\qquad\quad BE = DE = \dfrac{d_2}{2}$

Now consider $\triangle ABE$. Since, $\angle AEB$ is a right angle, hence using the formula for area of a right triangle, area of $\triangle ABE$ is equal to:

$\qquad\quad\begin{align} ar (\triangle ABE) &= \dfrac{1}{2}\dfrac{d_1}{2}\dfrac{d_2}{2}\\[4pt]
&= \dfrac{d_1 d_2}{8}\end{align}$

Here $ar$ stands for area. Similarly, area of other three triangles, i.e., $\triangle BCE, \triangle CDE$ and $\triangle DAE = \dfrac{d_1 d_2}{8}$.

$\begin{align}\qquad\quad\text{Area of ABCD} &= ar (\triangle ABE) + ar (\triangle BCE) + ar (\triangle CDE) + ar (\triangle DAE)\\
&= 4\times\dfrac{d_1 d_2}{8}\\
&= \dfrac{d_1 d_2}{2}\end{align}$

Q.E.D.

Recommended:
Area of triangle
Area of rectangle
Properties of rhombus

Square

noreply@blogger.com (Sampark Sathi) — Tue, 13 Jan 2015 12:39:00 +0000

Definition: Square is a quadrilateral having all the sides equal and all the interior angles as $90^o$. It may also be defined as a rectangle having equal side lengths or a rhombus having all the interior angles as $90^o$.

Area Of A Triangle

noreply@blogger.com (Sampark Sathi) — Mon, 12 Jan 2015 14:00:00 +0000

Theorem: Area of a triangle is equal to one half of the product of the length of any side and an altitude drawn over it by the vertex opposite to it.

Prerequisites:
Area of a right triangle (proof)

Proof:

In the given triangle ABC, draw an altitude from A to join BC at D. As a result, $\triangle ABC$ is divided into two right triangles, i.e., $\triangle ABD$ and $\triangle ACD$.

Since, these are right triangles, hence there area is given by:

$\qquad\quad\text{Area of $\triangle ABD$} = \dfrac{1}{2} (AD).(BD)\\
\text{And}\quad\text{Area of $\triangle ACD$} = \dfrac{1}{2}  (AD).(CD)$

$\begin{align}\therefore\;\;\quad\text{Area of $\triangle ABC$} &= \text{Area of $\triangle ABD$} + \text{Area of $\triangle ACD$}\\
&= \frac{1}{2}  (AD).(BD) + \frac{1}{2}  (AD).(CD)\\
&= \frac{1}{2}  (AD).(BD + CD)\\
&= \frac{1}{2}  (AD).(BC)\end{align}$

Hence the result.

Recommended:
Area of rectangle
Area of rhombus
Isosceles triangle theorem