Question

In figure given above, ABC is a right triangle right angled at C. If D is midpoint of BC, prove that AB² = 4AD² - 3AC²​

Answer 1

Given :  

∠C = 90° and D is the mid-point of BC.

In ∆ ACD,  

AD² = AC² + CD²  

[By using  Pythagoras theorem]

CD² = AD² - AC² ……….(1)

In ∆ACB,

AB² = AC² + BC²

[By using Pythagoras theorem]

AB² = AC² + (2CD)²

[D is the mid-point of BC]

AB² = AC² + 4CD²

AB² = AC² +4(AD² - AC²)

[from eq 1]

AB² = AC² + 4AD² - 4AC²

AB² = 4AD² - 4AC² + AC²

AB² = 4AD² - 3AC²

PLEASE MARK ME AS THE BRAINLIEST