Question

6. In A ABC, Z B = 90° and D is the mid-point of BC.
Prove that
(1) AC = ADP + 3CD2 (ii) BC = 4 (AD- AB?​

Answer 1

Step-by-step explanation:

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

Answer 2

Answer:

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

Step-by-step explanation:

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