Question

AOB is a diameter of the circle and C, D, E are any 3 points on the semicircle. Find the value of angle ACD +angle BED

Answer 1

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°

Answer 2

Answer:270

Step-by-step explanation:Join bc

Angle acb is equal to 90 (angle subtended by diameter)

Then quadrilateral bcde is cyclic quadrilateral

:. Angle bcd + Angle bed =180

Adding angle acb on both sides

i . e angle bcd+angle acb+angle bed=180+ angle acb

Angle acd+angle bed=180+90(angle bcd+angle acb=angle acd)(angle acb =90)

:. Angle acd + Angle bed=270

Hence proved

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