In △ABC, angle ∠ACB = 90 . D is a point on side AB so that DA = DC. (i) Prove that △BDC is an isosceles triangle. (ii) If ∠BDC = 60 , show that ∠A = 30 .

Question

KnowledgeBoat · Accepted Answer

(i) In △ADC, Since AD = DC, △ADC is an isosceles triangle. In isosceles triangle, angles opposite to equal sides are equal. ∴ ∠DAC = ∠DCA. Let ∠DCA be 'x'. In △ABC, ∠ACB = 90 ⇒ ∠ACB = ∠DCA + ∠DCB ⇒ ∠DCB = ∠ACB - ∠DCA ⇒ ∠DCB = 90 - x ........(1) By angle sum property of triangle, In triangle ABC, ⇒ ∠ACB + ∠A + ∠B = 180 ⇒ ∠B = 180 - (∠ACB + ∠A) ⇒ ∠B = 180 - (90 + x) ⇒ ∠B = 90 - x ........(2) From equation (1) & (2), we get : ∠DCB = ∠B Since, ∠DCB = ∠B, the sides opposite to them must be equal (CD = BD) ∴ △BDC is an isosceles triangle. Hence, proved that △BDC is an isosceles triangle. (ii) Given, ∠BDC = 60 From figure, ∠ADC + ∠BDC = 180 ∠ADC + 60 = 180 ∠ADC = 120 Since, triangle ADC is an isosceles triangle, with AD = CD. Thus, ∠DAC = ∠DCA = a (let) By angle sum property of triangle, ∠DAC + ∠DCA + ∠ADC = 180 a + a + 120 = 180 2a = 180 - 120 2a = 60 a = = 30 . Hence proved that ∠A = 30 .

Mathematics

In △ABC, angle ∠ACB = 90°. D is a point on side AB so that DA = DC.
(i) Prove that △BDC is an isosceles triangle.
(ii) If ∠BDC = 60°, show that ∠A = 30°.

Triangles

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