Question

In an isosceles triangle ABC;AB=AC and D is a point on BC prove that AB2 - AD2 = BD.CD

Answer 1

Here is the step by step explanation :

Draw the perpendicular from A to BC. Let it intersect BC at E.

Then in triangle AEB we have:

AB^2 = AE^2 + BE^2

In triangle AED we have:

AD^2 = AE^2 + BD^2

Subtracting gives:

AB^2 - AD^2 = BE^2 - BD^2

AB^2 - AD^2 = (BE + BD)(BE - BD)

Now, the perpendicular to the unequal side in an isosceles triangle is also the bisector of the unequal side. This can be proved by considering triangles AEB and AEC

AB = AC

angle AEB = angle AEC

AE = AE

By RHS criterion, the two triangles are congruent. So BE = CE

Depending on where you draw point E, you can get one of two things:

BE + ED = BD and BE - ED = CE - ED = CD

(or)

BE + ED = CE + ED = CD and BE - ED = BD

In either case you get: (BE+ED)*(BE-ED) = BD*CD.

Thus: AB^2 - AD^2 = BD*CD

Answer 2

Answer:

Step-by-step explanation:

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, therefore to find

In triangle, the value of AC is equal to

Therefore, using the different approach of AE i.e

We can say that

Hence, solving the value of

Now can also be written as according to the DE + CE = DE + BE.  

Therefore,

Hence, Proved