Question

In an isosceles triangle ABC; AB = AC and D is a point on BC produced. prove that :

AD2 = AC2 + BD.CD.

Answer 1

Mark E as midpoint of BC ,
hence AE is perpendicular to BC.

D lies on the extended BC. .

ED= 1/2 BC + CD
and, AE^2 = AC^2 - (1/2 BC) ^2. .

Now AD^2 = AE^2 + ED^2 . .

Substitute to get= AC^2 +CD^2 + BC.CD.
= AC^2 + BD.CD
as CD + BC = BD

Alternate Method :

if D is the midpoint of BC then by Pythagoras 

AC^2 = CD^2 + AD^2 

rearrange 

AD^2 = AC^2 - CD^2 

since D is a midpoint it would mean CD = BD, 

which means CD^2 is the same as CD.BD therefore 

AD^2 = AC^2 - CD.BD 

Answer 2

Answer:

Check your answer please