Question

ABC is a right angle to the triangle BAC. A point D is taken on BC that BD = AD. Proves that D is the midpoint of BC.​

Answer 1

Answer:

In ∆ACB and ∆ACD

Pythagoras thm 

AD²=DC²+AC² 

Hence, AC²=AD²-DC²________(1)

NOW,

AB²=AC²+BC²

AB²=(AD²-DC²)+BC²_____FROM(1)

AB²=AD²-(½BC)²+BC²______D is midpoint

AB²=AD²-¼BC²+BC²

Multipy by 4

4AB²=4AD²-DC²+4BC²

3AB²+AB²=4AD²+3BC²

AB²=4AD²+3BC²-3AB²

AB²=4AD²-3(BC²+AB²)

AB²=4AD²-3AC²_____Pythagoras thm

Hence Proved.

hopefully this will help you