Question

AB || DC in quadrilateral ABCD, ∠A = 76°, ∠BDC = 40° and ∠BCE = 125°. Prove that AD < DC.

Answer 1

Given,

AB || DC in quadrilateral ABCD, ∠A = 76°, ∠BDC = 40° and ∠BCE = 125°.

Exterior angle = Sum of interior angles

∠BCE = ∠BDC + ∠DBC

125° = 40° + ∠DBC

∠DBC = 85°

∠BCD + ∠BCE = 180°   ...(Linear pairs)

∠BCD + 125° = 180°

∠BCD = 55°

∠BDC = ∠ABD   ...(Alternate angles)

∠ABD = 40°

∠BDC = 40°

In ΔABD, ∠A + ∠ABD + ∠ADB = 180°

76° + 40° + ∠ADB = 180°

∠ADB = 64°

In ΔABD, ∠A > ∠ABD   ...(The side opposite to the largest angle is the longest in triangles)

BD > AD   ...(1)

In ΔBDC, ∠DBC > ∠BCD   ...(The side opposite to the largest angle is the longest in triangles)

DC > BD   ...(2)

From equation (1) and (2),

AD < DC

Hence, proved.