Question

Point D is the mid point of side BC of a right triangle ABC, right angled at c. prove that, 4AD²= 4 AC^2+ BC^2​

Answer 1

Answer:

Here's the answer look at the picture

Answer 2

Answer:

Given: ABC be a triangle, right - angled at C and D is the mid - point of BC.

To Prove: AB2 = 4AD2 - 3AC.

Proof:

From right triangle ACB, we have,

AB2 = AC2 + BC2

= AC2 + (2CD)2 = AC2 + 4CD2 [since BC = 2CD]

= AC2 + 4(AD2 - AC2) [From right △ACD]

= 4AD2 - 3AC2. 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.