1. ~~**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.

any help would be great

2. This situation?

3. 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

4. Originally Posted by Glaysher

Also angle BAC and angle AEC = 70 (by isosceles and circle theorem)
what is the circle theorum?

5. 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

6. 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