Question

$△ABC$ is an isosceles triangle in which $AB=AC$ and $D$ is a point on $BC$. then that relation is $A{B}^2-A{D}^2=BD\times CD$. ?

• A. True
• B. False

Answer 1

The correct option is A True

Due to being an isosceles triangle the sides $AB=AC$.

$AE$ is perpendicular to $BC$ as $AE$ is acting as a median for the triangle $ABC$ which is isosceles in nature.

Therefore, we can say that

$DE+CE=DE+BE$. Hence,$A{D}^2=A{E}^2+D{E}^2$$ therefore to find

$A{E}^2=A{D}^2-D{E}^2$

In triangle, the value of $AC$ is equal to $\sqrt{A{E}^2+E{C}^2}$

Therefore, using the different approach of $AE$ i.e

$A{E}^2=(A{D}^2-D{E}^{2}andA{C}^2-E{C}^2)$

We can say that $A{D}^2-D{E}^2=A{C}^2-E{C}^2$

Hence, solving the value of

$A{C}^2-A{D}^2=E{C}^2-D{E}^2$

we get

$D{E}^2-E{C}^2$ as $(DE+EC)(DE-EC)$

Now$(DE+EC)(DE-EC)$ can also be written as $CD\times BD$ according to the $DE+CE=DE+BE$.

Therefore,$A{C}^2-A{D}^2=CD\times BD$

Hence, Proved.