In a triangle ABC, AB = AC and ∠A = 36°. If the internal bisector of angle C meets AB at D, prove that AD = BC.

Given: In a triangle ABC, AB = AC and ∠A = 36°. The internal bisector of angle C meets AB at D. To prove: AD = BC Proof: Since AB = AC, Δ ABC is an isosceles triangle. ⇒ ∠B = ∠C Sum of all angles of the triangle is 180°. ⇒ ∠A + ∠B + ∠C = 180° ⇒ 36° + ∠B + ∠B = 180° ⇒ 36° + 2∠B = 180° ⇒ 2∠B = 180° - 36° ⇒ 2∠B = 144° ⇒ ∠B = ⇒ ∠B = ∠C = 72° Since CD is the angle bisector of ∠C, ⇒ ∠ACD = ∠BCD = x ∠C = x 72° = 36° In Δ ACD and Δ ABC, ∠DAC = ∠DCA = 36° AB = AC (Given) ∠CAB = ∠CAD (common angle) ⇒ AD = CD ...............(1) By ASA criterion, Δ ACD ≅ Δ ABC By CPCT, AD = CD. In Δ DCB, sum of all angles of the triangle is 180°. ⇒ ∠DCB + ∠DBC + ∠CDB = 180° ⇒ 36° + 72° + ∠CDB = 180° ⇒ 108° + ∠CDB = 180° ⇒ ∠CDB = 180° - 108° ⇒ ∠CDB = 72° As, ∠CDB = ∠CBD = 72° ⇒ CB = CD ..............(2) From (1) and (2), we get: AD = BC Hence, AD = BC.

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Mathematics

In a triangle ABC, AB = AC and ∠A = 36°. If the internal bisector of angle C meets AB at D, prove that AD = BC.

Triangles

Related Questions

The given figure shows a △ABC in which AB = AC and BP = CQ.
Prove that :
(i) △ABQ ≡ △ACP.
(ii) △APQ is isosceles.

Use the given figure to find the angle x.

In a triangle ABC, ∠A = x°, ∠B = (3x - 2)° and ∠C = y°. Also, ∠C - ∠B = 9°. Find all the three angles of the △ABC.

In the given figure, PR > PQ
Prove that : AR > AQ.

Mathematics

In a triangle ABC, AB = AC and ∠A = 36°. If the internal bisector of angle C meets AB at D, prove that AD = BC.

Triangles

Related Questions

The given figure shows a △ABC in which AB = AC and BP = CQ.Prove that :(i) △ABQ ≡ △ACP.(ii) △APQ is isosceles.

Use the given figure to find the angle x.

In a triangle ABC, ∠A = x°, ∠B = (3x - 2)° and ∠C = y°. Also, ∠C - ∠B = 9°. Find all the three angles of the △ABC.

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The given figure shows a △ABC in which AB = AC and BP = CQ.
Prove that :
(i) △ABQ ≡ △ACP.
(ii) △APQ is isosceles.

In the given figure, PR > PQ
Prove that : AR > AQ.