Question

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Answer 1

Given that : AD ⊥ BC and DB = 3CD

To prove : 2AB2 = 2AC2 + BC2

Proof :
BD + DC = BC

3CD + CD = BC

4CD = BC

CD = BC / 4

DB = 3CD = 3BC / 4.

In a right angle traingle ACD ,

AC2 = AD2 + CD2.

AC2 = AD2 + BC2 / 16 -------(1)

In a right angle traingle ABD ,

AB2 = AD2 + BD2.

AB2 = AD2 + 9BC2 / 16 -------(2).

Substracting (1) from (2) we obtain

AB2  - AC2 = 9BC2 / 16 - BC2 / 16

16(AB2  - AC2 ) = 8BC2

2(AB2  - AC2 ) = BC2

2AB2 = 2AC2 + BC2

Hence proved.

Answer 2

BC=DB+CD
     =3CD+CD
     =4CD
So, BC2=(4CD)2
               =16CD2
      CD2=BC2/16
In tri. ADB by pythagoras theorem,
AD2=AB2-BD2    ---i
In tri.ACD, by pythagoras theorem,
AD2=AC2-CD2    --ii
From i and ii, we have:
AB2-BD2=AC2-CD2
AB2-(3CD)2=AC2-CD2        (BD=3CD)
AB2-9CD2=AC2-CD2
AB2=AC2+8CD2
AB2=AC2+8(BC2/16)
AB2=AC2+BC2/2
Multiply both sides by 2
2AB2=2AC2+BC2          Proved