Need help to find Circle Question

[Hobbiez]

Hi Everyone,

 

Need some help to find the unknown angles BAD and CED.

Given angle BCD = 132. Angle A is at centre of circle.

 

Thanks for all input.

[Chisiang]

how about C? Center also?

[Chisiang]

is BCE a straight line?

[Hobbiez]

Yes angle C is also centre and BCE is straight line. Thanks for pointing that out.

Edited June 27, 2008 by Hobbiez

[David]

I think BAD = BCD = 132

[Chisiang]

CED is 66 deg irregardless of the radius of the circle

[Porker]

Angle CED = 66 degrees

 

Angle BAD = 132 degrees

[Hobbiez]

Thanks for your input. Yes I agree with you since BCE is stragiht line, we can find CED.

[Hobbiez]

Thanks David. May I know why BAD is equivalent to BCD?

[Hobbiez]

Thanks Porker. How come you change avatar again?

 

What is reason BAD is same as BCD?

[Porker]

Work with the following facts:

 

1) CED is isoceles

 

2) BCD is isoceles

 

3) BAD is isoceles

 

4) AD and BCE are parallel

[Skylander]

I dont know correct or not but here goes. Looks like Sec school question.

 

C = centre of circle, thus CD = CE = radius of circle

therefore triangle CDE is isosceles and angle CDE = angle CED.

Using theory of ext angle of triangle = sum of 2 int angle of triangle,

angle CED = 132/2 = 66 deg (since both angles CDE and CED are the same)

 

For angle BAD, using opposite angle in cyclic quad (is this possible?)

Angle BAD = 180 - 66 = 114 deg

 

Correct me if i am wrong

[Chisiang]

AD and BCE parallel meh? neber say

[Hobbiez]

Thanks Skylander. If I'm correct, cyclic Quad is when the four corners touch the same circle.

[Porker]

Some things no need to say can infer also. The examiners where got so nice give you everything?

[Hobbiez]

The question did not mention on this too. I apologised Chisiang.

[Skylander]

oh yah hor.... suddenly remember, thanks for pointing out. But at points ABDE, consider touch the same circle? Or must not pass through centre of circle?

[Hobbiez]

Hey you are good! Yes ABDE is a cyclic Quad.