In a circle with centre O, AB is a diameter. Points C, D and E are on the circle on one side of AB such that ABEDC is a pentagon. The sum of angles ACD and DEB is:

AOB is a diameter of the circle, C, E, D are three points on the circle

To find: ∠ACD + ∠BED

Construction: Join CO, DO and EO

Assume AC = CD = AO

Assume DE = EB

So in ΔACO

It is an equilateral triangle, hence ∠ACO = 60°

In ΔCDO

It is an equilateral triangle, hence ∠DCO = 60°

∠DOE = ∠EOB

As ∠AOB = 180°

So 180° = ∠AOC + ∠COD + ∠DOE + ∠EOB

And ∠AOC = ∠COD = 60°

And ∠DOE = ∠EOB

Hence 2∠DOE = 180° - 60° - 60°

∠DOE = 30°

In ΔODE

∠ODE = ∠OED (isosceles triangle)

And ∠ODE + ∠OED + ∠DOE = 180°

So 2∠ODE = 180° - 30° = 150°

So ∠ODE = 75°


In ΔOEB

∠OEB = ∠OBE (isosceles triangle)

And ∠OEB + ∠OEB + ∠BOE = 180°

So 2 ∠OEB = 180° - 30° = 150°

So ∠OEB = 75°



So ∠ACD + ∠BED = ∠ACO + ∠DCO + ∠DEO + ∠BEO = 60° + 60° + 75° + 75° = 270°