Question

Triangle ABC is right angled at B, and D is the mid- point of BC. Prove

that AC2 = AD2 + 3 BD2​

Answer 1

Given:

ABC is a right triangle, right-angled at B, and if D is the mid-point of BC.

To prove:

AC² = AD² + 3BD²

Solution:

For the given condition of the question

ABC is a right triangle, right angled at B, and if D is the mid-point of BC  

By the Pythagoras theorem

In ∆ABC

AB² + CB² =AC²

In ∆ADC

AB² + BD² =AD²

As we know that the D is the mid-point CB

So CD = BD

Replace this as per the need

AB² + (2BD)² =AC²

AB² + BD² =AD²

Subtract the above equations we get

4BD² – BD² = AC² - AD²  

AC² = AD² + 3BD²

Hence proved

Answer 2

Step-by-step explanation:

Given: Triangle ABC is right angled at B, and D is the mid- point of BC.

To prove:

AC² = AD² + 3BD²

Solution:

Alternative Method:

Apply Pythagoras theorem in ∆ABC

$AC^2=AB^2+BC^2... eq1$

Apply Pythagoras theorem in ∆ABD

$AD^2=AB^2+BD^2... eq2$

Subtract eq1-eq2

$AC^2-AD^2=AB^2+BC^2-AB^2-BD^2... eq2\\\\AC^2-AD^2=BC^2-BD^2... eq3$

since, D is midpoint of BC.

So,

BC= 2BD

Put this to eq3

$AC^2-AD^2=(2BD)^2-BD^2\\\\AC^2-AD^2=4BD^2-BD^2\\\\AC^2-AD^2=3BD^2\\\\\bold{AC^2=AD^2+3BD^2}$

Hence Proved.

Hope it helps you.

To learn more on brainly:

1)Triangle ABC is right angled at B, D is the midpoint of BC. Prove that, AC² = 4AD²– 3AB²

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2)In triangle ABC point D is midpoint of BC . Prove that AB2 = 4AD2 - 3AC2.

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