Q.14.In the trapezium ABCD(fig 2),BE is drawn ti cut DC at E such that AD=BE. Prove that angle ADE=angle EBC + angle BCE. Share with your friends Share 10 Varun.Rawat answered this Draw AL⊥CD and BM⊥CDIn ∆ALD and ∆BME∠ALD = ∠BME 90° each AD = BE Given AL = BM ⊥ distance between 2 ∥ lines is constant⇒∆ALD≅ ∆BME RHS⇒∠ADL = ∠BEM CPCT⇒∠ADE = ∠BEM .....1Now, ∠BEC + ∠BEM = 180° Linear pair⇒∠BEC = 180° - ∠BEM⇒∠BEC = 180° - ∠ADE Using 1In ∆BEC, ∠EBC + ∠BCE + ∠BEC = 180° Angle sum property⇒∠EBC + ∠BCE + 180° - ∠ADE = 180°⇒∠ADE = ∠EBC + ∠BCE 13 View Full Answer