Question

In an isosceles triangle ABC; AB = AC and D is the point on BC produced.

Prove that: AD² = AC² + BD.CD.

Answer 1

In an isosceles triangle ABC; AB = AC and

D is the point on BC produced. 

Construct AE perpendicular BC.

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

We consider the rt. angled ΔAED and applying Pythagoras theorem we get,

AD² = AE² + ED²

AD² = AE² + (EC + CD)²             ...(i) [∵ ED = EC + CD]

Similarly, in ΔAEC,

AC² = AE² + EC²

AE² = AC² - EC²                       ...(ii)

From equation (ii)

AD² = AC² - EC² + (EC + CD)²

AD² = AC² - EC² + EC² + CD² + 2EC·CD

AD² = AC² + CD (CD + 2EC)

AD² = AC² + CD (CD + BC)

AD² = AC² + CD·BD

Hence, proved.