Question

Triangle ABC is right angled at B, And D is the mid point of BC. Prove that AC² = 4AD² - 3AB²​

Answer 1

Answer:

GIVEN:

ABC is a right angled at B and D is the midpoint of BC.

In ∆ABD

AD²= AB²+BD²…………(1)

[By Pythagoras theorem]

In ∆ ABC

AC²= AB²+BC²…………(2)

From equation (1)

AD²= AB²+ (BC/2)²

[D is the midpoint of BC]

AD²= AB²+ BC²/4

AD²= AB²/1+ BC²/4

4AD²= 4AB²+ BC²

BC²= 4AD²-4AB²…………….(3)

Using the value of BC² in eq. 2

AC²=AB²+4AD²-4AB²

AC²=AB²-4AB+4AD²

AC²= 4AD²-3AB²

HENCE PROVED….

HOPE THIS WILL HELP YOU….

Answer 2

$\huge\underline\pink {Answer}$

$\large\bold\blue {Given}$

∆ABC with ∠B = 90°

D is the mid point of BC

$\large\bold\blue {To prove}$

AC² = 4AD² - 3AB²

$\large\bold\blue {Proof}$

In ∆ABC,

∠B = 90°

⇒ AC² = AB² + BC² (By Pythagoras theorem)

⇒ = AB²+(2BD)²

⇒ AC² = AB²+ 4BD² ⇢ (i)

In ∆ABD, AD² = AB² + BD² (By Pythagoras theorem)

⇒ BD² = AD² - AB² ⇢ (ii)

From (i) and (ii) we get

AC² = AB² + 4(AD²-AB²)

= AB² + 4AD - 4AB²

AC² = 4AD² - 3AB²

$\huge\fbox\fcolorbox{pink}{red} {Hence proved}$