Q) In Δ ABC, if AD ⊥ BC and AD² = BD x DC, then prove that ∠BAC = 90⁰

Ans:

Since AD ⊥ BC, ∴ ∠ ADB = ∠ ADC = 90⁰

By Pythagoras theorem in Δ ADB, we get:

AD² + BD² = AB² ……………. (i)

Similarly in Δ ADC, by Pythagoras theorem, we get:

AD² + DC² = AC² ……………..(ii)

By adding both equations (i) and (ii), we get:

(AD² + BD²) + (AD² + DC²) = AB² + AC²

2 AD² + BD² + DC² = AB² + AC²

Given that AD² = BD x DC

Therefore, the above equation will change to:

(BD x DC) + BD² + DC² = AB² + AC²

Since a² + b² + 2 a b = (a + b)²

∴ (BD + DC)² = AB² + AC²

Since BD + DC = BC

∴ BC² = AB² + AC²

∴ This is possible if Δ ABC is a right-angled triangle

Therefore ∠ BAC = 90⁰

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