Question

In ABC, ZB = 90° and D is the mid-point of BC. Prove that AC^2 = AD^2 + 3CD^2.
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Answer 1

Given: In △ABC, ∠B = 90° and D is the mid-point of BC.

To Prove: AC2 = AD2 + 3CD2

Proof:

In △ABD,

AD2 = AB2 + BD2

AB2 = AD2 - BD2 .......(i)

In △ABC,

AC2 = AB2 + BC2

AB2 = AC2- BD2 ........(ii)

Equating (i) and (ii)

AD2 - BD2 = AC2 - BC2

AD2 - BD2 = AC2 - (BD + DC)2

AD2 - BD2 = AC2 - BD2- DC2- 2BDx DC

AD2 = AC2 - DC2 - 2DC2 (DC = BD)

AD2 = AC2 - 3DC2.

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Answer 2

Given: In △ABC, ∠B = 90° and D is the mid-point of BC.

To Prove: AC2 = AD2 + 3CD2

Proof:

In △ABD,

AD²= AB² + BD²

AB²= AD² - BD².......(i)

In △ABC,

AC²= AB² + BC²

AB² = AC²- BD² ........(ii)

Equating (i) and (ii)

AD²- BD² = AC² - BC²

AD²- BD2 = AC2 - (BD + DC)²

AD²- BD²= AC²- BD²- DC²- 2BDx DC

AD² = AC² - DC² - 2DC² (DC = BD)

AD² = AC²- 3DC²

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