Geometry Problem

Post your solution in the comment box below

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 925.

## Friday, September 20, 2013

### Geometry Problem 925: Cyclic Quadrilateral, Diagonal, Circle, Angle Bisector, Parallel

Labels:
angle bisector,
circle,
cyclic quadrilateral,
diagonal,
parallel

Subscribe to:
Post Comments (Atom)

http://img39.imageshack.us/img39/8260/xhe.bmp

ReplyDeleteLet EG cut CF at K. define angles x and y as shown on the sketch

In triangle BEK we have y=∠ (BEG)- ∠ (CBD)=1/4(Arc(AB)+Arc(CD))- 1/2Arc(CD)= 1/4(Arc(AB)-Arc(CD))

We have x= ½∠ (BFA)=1/2 x ½( Arc(AB)-Arc(CD))= 1/4(Arc(AB)-Arc(CD))

So x=y and EG//FH

Let EG and FH cut AB at P,Q respectively

ReplyDelete∠BFQ = 1/2∠BFA = 1/2(90 - A/2 - B/2)

∠BQF = a + ∠BFQ = 90 + A/2 - B/2

Now let ∠CBD = x, ∠CDB = y, so that x+y=a.

∠PEA = 1/2∠BEA = 1/2∠(180 - y - (B-x)) = 90 - y/2 - B/2 + x/2

∠BPE = ∠PEA + y = 90 - B/2 + x/2 + y/2 = 90 - B/2 + A/2

So ∠BPE=∠BQF

q.e.d.

See following picture:

ReplyDeletehttp://img40.imageshack.us/img40/9981/ptg.gif