Question

in right angled ∆ ABC in which angle C=90 , if D is the mid point of BC , prove that AB^2 = 4AD^2-3AC^2​

Answer 1

Explanation :

Given :

• In ∆ABC, ∠C = 90°
• D is the mid point of BC.

To prove :

• AB² = 4AD² - 3AC²

proof :

In ∆ACB, using Pythagoras theorem,

AB² = AC² + BC² ..........(1)

Now, it is given that, D is the mid point of BC,

•°• BD = DC

Put BC = DC in (1)

=> AB² = AC² + (2CD)²

=> AB² = AC² + 4CD².........(2)

In ∆ACD, by using Pythagoras theorem,

AD² = AC² + CD²

=> CD² = AD² - AC².........(3)

Put the value of (3) in (2),

=> AB² = AC² + 4(AD² - AC²)

=> AB² = AC² + 4AD² - 4AC²

=> AB² = 4AD² - 3AC²

Hence proved!

Answer 2

GIVEN,

<C=90°

BD=DC=1/2BC.......(i)

To Prove,

Proof,

In ΔABC,

By Pythagoras Theorem,

AB^2=BC^2+AC^2........(ii)

Now,

Similarly,By Pythagoras theorem,

In ΔADC,

AD^2=AC^2+DC^2

=>AD^2–DC^2=AC^2.....(iii)

From (ii) and (iii),

AB^2=BC^2+AD^2–DC^2

=>AB^2=BC^2+AD^2–(BC/2)^2 [From iii]

=>AB^2=BC^2+AD^2–(BC^2)/4 [From ii]

=>4AB^2=4AD^2+3AB^2–3AC^2

=>AB^2=4AD^2–3AC^2

Hence,Proved.