#### Question

In the figure, given below, AD = BC, ∠BAC = 30° and ∠CBD = 70°.

Find: (i) ∠BCD (ii) ∠BCA (iii) ∠ABC (iv) ∠ADB

#### Solution

In the figure, ABCD is a cyclic quadrilateral

AC and BD are its diagonals

∠BAC = 30° and ∠CBD = 70°

Now we have to find the measure of

∠BCD, ∠BCA, ∠ABC and ∠ADB

We have ∠CAD = ∠CBD = 70° [Angles in the same segment] Similarly, ∠BAD = ∠BDC = 30°

∴ ∠BAD = ∠BAC + ∠CAD

= 30° + 70°

= 100°

(i) Now ∠BCD + ∠BAD =180° [Opposite angles of cyclic quadrilateral]

⇒ ∠BCD + ∠BAD = 180°

⇒ ∠BCD + 100° = 180°

⇒ ∠BCD = 180° -100°

⇒ ∠BCD = 80°

(ii) Since AD = BC, ABCD is an isosceles trapezium and AB ∥ DC

∠BAC = ∠DCA [Alternate angles]

⇒ ∠DCA = 30°

∴ ∠ABD = ∠DAC = 30° [Angles in the same segment]

∴ ∠BCA = ∠BCD - ∠DAC

= 80° - 30°

= 50°

(iii) ∠ABC = ∠ABD + ∠CBD

= 30° + 70°

= 100°

(iv) ∠ADB = ∠BCA = 50° [Angles in the same segment]