# ~~**circle geometry**~~

• Aug 22nd 2006, 02:31 AM
fring
~~**circle geometry**~~
in a triangle ABC, AC=BC and angle ACB=40. The vertices A,B and C lie on the circumference of a cirlce. D is a point outside this circle such that DC is perpendicular to BC, BC=DC and the points B and D are on the same side of the line AC. DA cuts the circumference of the circle at E. Find the size of angle BCE.

:confused: :confused: :confused:

any help would be great
• Aug 22nd 2006, 03:11 AM
Glaysher
This situation?
• Aug 22nd 2006, 03:22 AM
Glaysher
ABC isosceles

So angle ABC = $\displaystyle \frac{180-40}{2}=70$

Also angle BAC and angle AEC = 70 (by isosceles and circle theorem)

ACD isosceles

So angle CAE = $\displaystyle \frac{180-40-90}{2}=25$

Call unmarked point P where lines cross

angle APC = 180 - 25 - 40 = 115

So angle CPE = 65

So angle BCE = 180 - 65 - 70 = 45
• Aug 22nd 2006, 06:00 AM
Quick
Quote:

Originally Posted by Glaysher

Also angle BAC and angle AEC = 70 (by isosceles and circle theorem)

what is the circle theorum?
• Aug 22nd 2006, 06:09 AM
Glaysher
There are lots of circle theorems. This is a link for them. I used the first one.

http://www.mathsrevision.net/gcse/pages.php?page=13
• Aug 22nd 2006, 06:12 AM
Quick
Quote:

Originally Posted by Glaysher
There are lots of circle theorems. This is a link for them. I used the first one.

http://www.mathsrevision.net/gcse/pages.php?page=13

Thanks