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Questions
In triangle ABC, AD is perpendicular to side BC and AD2 = BD × DC. Show that angle BAC = 90°.
In ΔABC, if AD⊥BC and AD2 = BD × DC, then prove that ∠BAC = 90°.
Solution
Given: AD2 = BD × DC
`(AD)/(DC) = (BD)/(AD)`
∠ADB = ∠ADC = 90°
∴ ∆DBA ~ ∆DAC ...(SAS similarity)
So, these two triangles will be equiangular.
∴ ∠1 = ∠C and ∠2 = ∠B
∠1 + ∠2 = ∠B + ∠C
∠A = ∠B + ∠C ...(i)
By angle sum property,
∠A + ∠B + ∠C = 180°
∠A + ∠A = 180° ...[From (i) we get]
2∠A = 180°
∠A = ∠BAC = 90°
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