Question

ABC is a right triangle, right angled at A and D is mid point of AB. Prove that, BC²=CD²+3BD²​

Answer 1

Answer:

In triangle ABC

By pythagoras theorem

BC²=AB²+AC²....(1)

In triangle ADC

CD²=AC²+AD²

AC²=CD²-AD²

from (1)  

BC²=AB²+CD²-AD²

BC²=(AD+BD)²+CD²-AD²

BC²=AD²+2AD.BD+BD2+CD²-AD²

BC²=2AD.BD+BD²+CD²

since D is a midpoint of AB

AD=BD

BC²=2BD²+BD²+CD²

BC²=3BD²+CD²

Hence Proved.....