**Q) In Δ ABC, if AD ⊥ BC and AD ^{2} = BD x DC, then prove that ∠BAC = 90^{0}**

**A****ns:**

Since AD ⊥ BC, ∴ ∠ ADB = ∠ ADC = 90^{0}

By Pythagoras theorem in Δ ADB, we get:

AD^{2} + BD^{2} = AB^{2} ……………. (i)

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

AD^{2} + DC^{2} = AC^{2} ……………..(ii)

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

(AD^{2} + BD^{2}) + (AD^{2} + DC^{2}) = AB^{2} + AC^{2}

2 AD^{2} + BD^{2} + DC^{2} = AB^{2} + AC^{2}

Given that AD^{2} = BD x DC

Therefore, the above equation will change to:

(BD x DC) + BD^{2} + DC^{2} = AB^{2} + AC^{2}

Since a^{2} + b^{2} + 2 a b = (a + b)^{2}

∴ (BD + DC)^{2} = AB^{2} + AC^{2}

Since BD + DC = BC

∴ BC^{2} = AB^{2} + AC^{2}

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

**Therefore ∠ BAC = 90 ^{0}**

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