Name: Illustration Based on the SAS Criterion
Uploaded: 2019-07-19
Duration: 4 min 36 s
Description: One of the easiest ways to learn a concept is through a visual illustration. To learn how the SAS congruence criterion is applied in mathematical problems, w...

Question

$A B$ is the hypotenuse of right isosceles triangle $A B C$ . $A D$ is the bisector of angle $∠ B A C$ and $A D$ intersects $B C$ at $D$ . Let's prove that $A C + C D = A B$

Accepted Answer

Let $A B C$

In a right-angled triangle $B C A$ ,

By Pythagoras theorem,

$A B^{2} = B C^{2} + A C^{2}$

$A B^{2} = a^{2} + a^{2}$

$A B = a \sqrt{2}$

By using bisector theorem,

$\frac{A C}{A B} = \frac{C D}{B D} \Rightarrow \frac{A B}{A C} = \frac{B D}{C D}$

Adding $1$ on both sides, we get

$\frac{A B}{A C} + 1 = \frac{B D}{C D} + 1$

$\frac{A B + A C}{A C} = \frac{B D + C D}{C D}$

$\frac{A B + A C}{A C} = \frac{B C}{C D}$

$\frac{a \sqrt{2} + a}{a} = \frac{a}{C D}$

Now, $C D = \frac{a}{\sqrt{2} + 1}$

So, $A C + C D = a + \frac{a}{\sqrt{2} + 1}$

$= \frac{\sqrt{2} a + a + a}{\sqrt{2} + 1} = \frac{\sqrt{2} a + 2 a}{\sqrt{2} + 1} = \frac{\sqrt{2} a (1 + \sqrt{2})}{\sqrt{2} + 1} = \sqrt{2} a = A B$

Hence, $A C + C D = A B$

$A B$ is the hypotenuse of right isosceles triangle $A B C$ . $A D$ is the bisector of angle $∠ B A C$ and $A D$ intersects $B C$ at $D$ . Let's prove that $A C + C D = A B$

Important Questions on Relation Between Two Sides of a Triangle and Their Opposite Angles

Learn and Prepare for any exam you want !

AB is the hypotenuse of right isosceles triangle ABC. AD is the bisector of angle ∠BAC and AD intersects BC at D. Let's prove that AC+CD=AB

Important Questions on Relation Between Two Sides of a Triangle and Their Opposite Angles

Learn and Prepare for any exam you want !

$A B$ is the hypotenuse of right isosceles triangle $A B C$ . $A D$ is the bisector of angle $∠ B A C$ and $A D$ intersects $B C$ at $D$ . Let's prove that $A C + C D = A B$