The perpendicular AD on the base BC of a ∆ABC intersects BC at D so that DB = 3 CD. Prove that `2"AB"^2 = 2"AC"^2 + "BC"^2`

#### Solution 1

We have

DB = 3CD

BC = BD + DC

The perpendicular AD on the base BC of a ∆ABC intersects BC at D so that DB = 3 CD. Prove that 2AC2 + BC2.

We have,

DB = 3CD

∴ BC = BD + DC

⇒ BC = 3 CD + CD

`⇒ BD = 4 CD ⇒ CD = \frac { 1 }{ 4 } BC`

`∴ CD = \frac { 1 }{ 4 } BC and BD = 3CD = \frac { 1 }{ 4 } BC ….(i)`

Since ∆ABD is a right triangle right-angled at D.

`∴ AB^2 = AD^2 + BD^2 ….(ii)`

Similarly, ∆ACD is a right triangle right angled at D.

`∴ AC^2 = AD^2 + CD^2 ….(iii)`

Subtracting equation (iii) from equation (ii) we get

`AB^2 – AC^2 = BD^2 – CD^2`

`⇒ AB^2 – AC^2 = ( \frac{3}{4}BC)^{2}-( \frac{1}{4}BC)^{2}[`

`⇒ AB^2 – AC^2 = \frac { 9 }{ 16 } BC^2 – \frac { 1 }{ 16 } BC^2`

`⇒ AB^2 – AC^2 = \frac { 1 }{ 2 } BC^2`

`⇒ 2(AB^2 – AC^2 ) = BC^2`

`⇒ 2AB^2 = 2AC^2 + BC^2`

#### Solution 2

In ΔACD

AC^{2} = AD^{2 }+ DC^{2}

AD^{2} = AC^{2} - DC^{2} ...(1)

In ΔABD

AB^{2 }= AD^{2 }+ DB^{2}

AD^{2} = AB^{2} - DB^{2} ...(2)

From equation (1) and (2)

Therefore AC^{2} - DC^{2} = AB^{2} - DB^{2}

since given that 3DC = DB

DC = `"BC"/(4) and "DB" = (3"BC")/(4)`

`"AC"^2 - ("BC"/4)^2 = "AB"^2 - ((3"BC")/4)^2`

`"AC"^2 - "Bc"^2/(16) = "AB"^2 - (9"BC"^2)/(16)`

16AC^{2 }- BC^{2 }= 16AB^{2} - 9BC^{2}

⇒ 16AB^{2} - 16AC^{2 }= 8BC^{2}

⇒ 2AB^{2 }= 2AC^{2} + BC^{2}.