Question

In ∆ABC, ∠ABC = 135°. Prove that AC² = AB² + BC² + 4 ar (∆ABC)

Answer 1

We have the following figure.

Let Draw an altitude AD on Extended BC

In ∆ADC, ∠D = 90°

Therefore, by Pythagoras theorem, we have

`AC^2=AD^2+DC^2`

`AC^2=AD^2+(DB+BC)^2`

`AC^2=AD^2+DB^2+BC^2+2.BC.BD`   ...(i)

In ΔADB, ∠D = 90°

⇒ `AD^2 + BD^2 = AB^2`

By substituting values of equation (i)

⇒`AD^2+DB^2+BC^2+2.BC.BD = AC^2`

⇒`AB^2 + BC^2 + 2BD.BC = AC^2`

As in ΔADB, ∠DAB = 45°

⇒ AD = DB (Opposite sides are equal)

⇒`AB^2+BC^2+2.BC.BD = AC^2`

⇒`AB^2+BC^2+2.BC.AD = AC^2`

⇒`AB^2+BC^2 + 4 xx 2/4 xx AD xx BC= AC^2`

∴ `1/2 xx AD xx BC = ar(ΔABC)`

⇒ `AC^2 = AB^2 + BC^2 + 4 ar (∆ABC)`