Question

In $\Delta ABC$, AB = AC. Side BC is produced to D. Prove that $(A{D}^2-A{C}^2)=BD.CD$.

Answer 1

Const: Draw a perpendicular AE from A

Thus, AE ⊥BC

Proof:

In ΔABC, AB = AC

And AE is a bisector of BC

Then, BE = EC

In right angle triangles ΔAED and ΔACE

$A{D}^2=A{E}^2+D{E}^2$ —— (1)

$A{C}^2=A{E}^2+C{E}^2$ ——- (2)

Subtracting (2) from (1),

⇒$(A{D}^2-A{C}^2)=D{E}^2-C{E}^2$

= $A{D}^2-A{C}^2=(DE+CE)(DE-CE)=(DE+BE)(DE-CE)$ [since CE = BE]

⇒$A{D}^2-A{C}^2=BD\times CD$

Hence proved