## Sunday, October 26, 2008

### Elearn Geometry Problem 198: Triangle, Quadrilateral, Angle

See complete Problem 198 at:

Triangle, Quadrilateral, Circle, Angles. Level: High School, SAT Prep, College geometry

1. maps D to D' by reflecting about BC
B is the circumcenter of ADD', ACD' is a straight line and CDD' is an isos. triangle
x=2mCD'D=50°

2. better solution:
maps D to D' by reflecting about BC
ACD' is a straight line, BAD' is an isos. triangle
mCAB=mBD'C=mCDB, and hence x=50°

3. About Jankonyex's solution for problem 198.
Could anybody tell me why is ACD' a straight line?
I think that this explanation is missing.
Thanks.

1. Nilton

See sketch below for detail.
http://img692.imageshack.us/img692/6971/problem198c.png
Peter Tran

4. To Peter, about problem 198.
Thank you for your help. Now I understand the solution by Jankonyex.
I wish to propose other solution, I hope it's right.

Let O be the circumcenter of triangle ACD. The straight line BO, which is the line bisector of AD, cuts that circle at P, that is midpoint of the arc AD. So CP is angle bisector of ACD. Then we have ang(BCP) = 65º + 25º = 90º.
The circle O is also circumscribed to CDP, so the perpendicular OQ to CP is line bisector of CP and CQ = QP.
But OQ is parallel to BC so O is midpoint of BP. This means that B belongs to the circle O
and ABCD is cyclic. Thus x = ang(ABD) = ang(ACD) = 50º.

5. Draw altitudes BP and BQ to DC and AC respectively which are easily seen to be equal since BC bisects < ACP.
As a result right Tr.s ABQ and DBP are congruent from which it follows that < BAC = < BDC. Hence ABCD is cyclic and x = 50.