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In triangle ABC, AD is perpendicular to side BC and AD2 = BD × DC.

Show that angle BAC = 90°.

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Answer

Triangle ABC is shown in the figure below:

In triangle ABC, AD is perpendicular to side BC and AD^2 = BD × DC. Show that angle BAC = 90°. Similarity, Concise Mathematics Solutions ICSE Class 10.

Given :

AD2 = BD × DC

⇒ ADDC=BDAD\dfrac{AD}{DC} = \dfrac{BD}{AD}

∠ADB = ∠ADC [Both = 90°]

∴ △DBA ~ △DAC (By SAS).

Since, triangles are similar they will be equiangular.

∴ ∠1 = ∠C and ∠2 = ∠B

⇒ ∠1 + ∠2 = ∠B + ∠C

⇒ ∠A = ∠B + ∠C

By angle sum property :

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠A = 180°

⇒ 2∠A = 180°

⇒ ∠A = 90°.

From figure,

⇒ ∠BAC = ∠A = 90°.

Hence, proved that ∠BAC = 90°.

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