is the altitude of Δ ABC such that B—D—C. If AD² = BD ⋅ DC, prove that ∠BAC is 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 AD² = BD.DC …… (3)

From equation (1) and (3), we get,

AB² = BD² + BD.DC

⇒ AB² = BD(BD + DC) = BD. BC

Now, from equation (2) and (3),

AC² = BD.DC + DC²

⇒ AC² = DC (BD + DC)

⇒ AC² = DC(BC)

Also, BC = BD + DC,

Substituting the value of BD and DC in above equation, we get,

⇒ BC² = AB² + AC²

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