Question

In the given figure, AD bisects ∠BAC and DE bisects ∠BDA. Find ∠BED.

Answer 1

∠BAC = 180° – (40° + 80°)

∠BAC = 60°

AB bisects ∠A

∠BAD = ∠DAC

= `60/2`

= 30°

∠ABD + ∠BDA + ∠DAB = 180°

40° + ∠BDA + 30° = 180°

∠BDA = 180° – 70°

∠BDA = 110°

DE bisects ∠BDA

∠BDE = ∠EDA

= `110/2`

= 55°

Now, In ΔBED

∠EBD + ∠BDE + ∠DEB = 180°

40° + 55° + ∠DEB = 180°

∠DEB = 180° – 95°

∠DEB = 85°