Question

In triangle ABC, angleB=90° and D is the midpoint of sideBC. Prove that:-
$ac { }^{2} = ad {}^{2} + 3cd {}^{2}$


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Answer 1

Answer:

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

To Prove: AC^2 = AD^2 + 3CD^2

Proof:

In △ABD,

AD^2 = AB^2 + BD^2

AB^2 = AD^2 - BD^2.......(i)

In △ABC,

AC^2 = AB^2 + BC^2

AB^2 = AC^2- BD^2 ........(ii)

Equating (i) and (ii)

AD^2 - BD^2 = AC^2- BC^2

AD^2 - BD^2 = AC^2 - (BD + DC)^2

AD^2 - BD^2 = AC^2 - BD^2- DC^2- 2BDx DC

AD^2 = AC^2 - DC^2 - 2DC^2 (DC = BD)

AD^2= AC^2 - 3DC^2

AC^2 = AD^2+3CD^2

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