In ΔABC, <img wid

Since AD is the altitude to BC,

⇒ ∠BDA = ∠CDA = 90°

So, Δ ABD and Δ ACD are right angle triangles.

Using Pythagoras theorem,

AB² = BD² + AD² ...... (1)

And, AC² = AD² + DC² ...... (2)

It is given that AC² = CD.BC

From equation (1) and (2), we get,

AB² – BD² = AC² – CD²

⇒ AB² – (BC – CD)² = AC² – CD²

⇒ AB² – BC² – CD² + 2.BC.CD = AC² – CD²

⇒ AB² – BC² + 2.BC.CD = AC²

Substituting the value of CD above, we get,

⇒ AB² – BC² + 2AC² = AC²

⇒ AB² + AC² = BC²

So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.