In the give figure, ABCD is a cyclic quadrilateral in which BC = CD and ∠CBD = 35°. Then, ∠BAD = ?
(a) 65°
(b) 70°
(c) 110°
(d) 90°

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Solution

(b) 70°
BC = CD (given)
⇒ ∠BDC = ∠CBD = 35°
In Δ BCD, we have:
∠BCD + BDC + ∠CBD = 180° (Angle sum property of a triangle)
⇒ ∠BCD + 35° + 35° = 180°
⇒ ∠BCD = (180° - 70°) = 110°
In cyclic quadrilateral ABCD, we have:
∠BAD + ∠BCD = 180°
⇒ ∠BAD + 110° = 180°
∴ ∠BAD = (180° - 110°) = 70°

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Similar questions

In the given figure, ABCD is a cylic quadrilateral in which BC = CD and $∠ C B D = 35^{\circ} .$ Then, $∠ B A D = ?$

(a) $65^{\circ}$

(b) $70^{\circ}$

(d) $90^{\circ}$

A B C D

A D = B C

∠ A = 110^{\circ}

∠ B

∠ D B C = 55^{\circ}

∠ B A C = 45^{\circ}

∠ B C D

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