In ΔABC, , B—D—C. If AB² = BD ⋅ BC, prove that ∠ BAC is a right angle.

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 AB² = BD.BC

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

AB² – BD² = AC² – DC²

⇒ AB² – BD² = AC² – (BC – BD)²

⇒ BC² + BD² – 2BC.BD = AC² + BD² – AB²

⇒ BC² – 2BC (BD) = AC² – AB²

Substituting the value of BD above, we get,

⇒ BC² – 2(AB)² = AC² – AB²

⇒ BC² = AB² + AC²

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