In triangle ABC, AD is perpendicular to side BC and AD² = BD × CD.
prove that angle BAC = 90°.

Question

In triangle ABC, AD is perpendicular to side BC and AD² = BD × CD. Prove that angle BAC = 90°.

KnowledgeBoat · Accepted Answer

Triangle ABC is shown in the figure below: Given : AD2 = BD × CD ⇒ ∠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°.

Mathematics

In triangle ABC, AD is perpendicular to side BC and AD² = BD × CD.
prove that angle BAC = 90°.

Similarity

Related Questions

In the given figure, Δ ABC is isosceles and AP x BQ = AC², prove that Δ ACP ∼ Δ BCQ.

Mathematics

In triangle ABC, AD is perpendicular to side BC and AD2 = BD × CD.prove that angle BAC = 90°.

Similarity

Related Questions

In the given figure, Δ ABC is isosceles and AP x BQ = AC2, prove that Δ ACP ∼ Δ BCQ.

In triangle ABC, AD is perpendicular to side BC and AD² = BD × CD.
prove that angle BAC = 90°.

In the given figure, Δ ABC is isosceles and AP x BQ = AC², prove that Δ ACP ∼ Δ BCQ.