Question

In ΔАВС, AB = AC and D is any point on side BC produced. Prove that: AD² = AB² + BD · CD.

Answer 1

Given:

In triangle ABC, AB = AC.

D is any point on BC produced i.e., B, C, D are collinear.

To Prove:

AD² = AB² + BD · CD.

Proof (Step-wise):

1. Place a convenient coordinate system:

Put BC on the x-axis, with B = (–b, 0) and C = (b, 0).

2. Since AB = AC, A lies on the perpendicular bisector of BC; take A = (0, h) for some h.

3. Let D be any point on the line BC (produced); write D = (d, 0) for some real d.

4. Compute squared distances:

AD² = (d – 0)² + (0 – h)² 

= d² + h²

AB² = (–b – 0)² + (0 – h)² 

= b² + h²

5. Compute BD and CD and their product:

BD = Distance from B to D

= d – (–b)

= d + b

CD = Distance from C to D

= d – b

BD × CD

= (d + b)(d – b)

= d² – b²

6. Now combine AB² and BD · CD:

AB² + BD · CD

= (b² + h²) + (d² – b²) 

= d² + h²

= AD²

7. Therefore, AD² = AB² + BD · CD, as required.