Question

In an isosceles triangle ABC;AB=AC and D is a point on BC produced. Prove that: AD^2=AC^2+BD×CD

Answer 1

Solution:

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 $A E^{2}=\left(A D^{2}-D E^{2} \text { and } A C^{2}-E C^{2}\right)$

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

Hence, solving the value of $A D^{2}-A C^{2}=D E^{2}-E C^{2} we get D E^{2}-E C^{2} \text { as }(D E-E C)(D E+E C)$

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

Therefore, $\bold{A D^{2}-A C^{2}=C D \times B D.}$

Hence, Proved.

Answer 2

Answer:

hope that will help

Step-by-step explanation:

follow the pictures