Question

In ΔABC, AD is perpendicular to BC and BD : DC = 3 : 1. Prove that `AB^2 = AC^2 + 1/2 BC^2`.

Answer 1

Using Pythagoras theorem,

In ΔADB,

AD² + DB² = AB²   ...(1)

In ΔADC,

AD² + DC² = AC²   ...(2)

Subtracting the equations we have,

⇒ AD² + DB² – AD² – DC² = AB² – AC²

⇒ (3CD)² – DC² = AB² – AC²

⇒ 9CD² – DC² = AB² – AC²

⇒ 8CD² = AB² – AC²

But BC = BD + CD

⇒ 3CD + CD = 4CD

⇒ CD = `(BC)/4`

⇒ `8 xx (1/4 BC)^2 + AC^2 = AB^2`

⇒ `8 xx 1/16 BC^2 + AC^2 = AB^2`

⇒ `1/2 BC^2 + AC^2 = AB^2`

Hence, proved.