Question

In an isosceles $\Delta$ ABC, AB = AC and D is a point on BC produced. Prove : $A{D}^2=A{C}^2+BD.CD$

Answer 1

Given: $△ABC$, $AB=AC$, D is a point on BC produced
Construction: Draw $AM\perp BC$
Since ABC is an Isosceles triangle and $AM\perp BC$, AM bisects BC
Thus, $BM=CM$
Now, In $△ACM$
$A{C}^2=A{M}^2+C{M}^2$ (Pythagoras Theorem)
$A{M}^2=A{C}^2-C{M}^2$ (I)
Now, In $△AMD$,
$A{D}^2=A{M}^2+M{D}^2$ (Pythagoras theorem)
$A{M}^2=A{D}^2-M{D}^2$ (II)
Equating I and II
$A{C}^2-C{M}^2=A{D}^2-M{D}^2$
$A{D}^2=A{C}^2+M{D}^2-C{M}^2$
$A{D}^2=A{C}^2+(MD-CM)(MD+CM)$
$A{D}^2=A{C}^2+\left(CD\right)(MD+BM)$ (CM = BM)
$A{D}^2=A{C}^2+CD\times BD$