Question

In triangle ABC, angle A = 90^o, CA = AB and D is the point on AB produced.
Prove that DC² - BD² = 2AB.AD.

Answer 1

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  ΔACD and applying Pythagoras theorem we get,
CD² = AC² + AD²
CD² = AC² + ( AB + BD )²                    ....[ ∵ AD = AB + BD ]
CD² = AC² + AB² + BD² + 2AB.BD      ...(i)

Similarly, in ΔABC,
BC² = AC² + AB²
BC² = 2AB²                                        ...[ AB = AC ]
AB² = `1/2`BC²                                   ...(ii)

Putting, AB2 from (ii) in (i), We get,
CD² = AC² + `1/2`BC² + BD² + 2AB . BD

CD² - BD² = AB² + AB² + 2AB . ( AD - AB )

CD² - BD² = AB² + AB² + 2AB . AD - 2AB² 

CD² - BD² = 2AB . AD

DC² - BD² = 2AB . AD                       

Hence Proved.