## Thursday, June 28, 2012

### Problem 774: Triangle, Two Altitudes, Circumcircle, Tangent at a Vertex, Parallel

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

Click the figure below to see the complete problem 774.

1. Note that angle FBD = angle BAC
Since A,E,D,C are concyclic, we have angle BAC = angle EAC = angle BDE

Hence angle FBD = angle BDE

So BF is parrallel to DE

q.e.d.

2. Let be P the intersection of rays AD and BF. Note two thing:
(1) <BDP=90°
(2) <CAB=<CBP.
Now, in triangle AEC and BPD we have <CAE + <ECA = 90° = <CBP+<BPD. Then <ECA=<BPD. As AEDC is cyclic, <EDA=<ECA=<BPD.

Therefore BF is parallel to BP.

Greetings go-solvers.

1. Note that, if we call P and Q to the intersections of the rays AD and CE with the line BF, the quadrilateral AQPB, is cyclic. Nice... ^^

3. BF(or any line parallel to BF like ED)is said to be antiparallel to AC.
The locus of the midpoints of all anti parallel segments (eah parallel to BF)is a straight line. Suppose this straight line meets AC at Y. We call BY a symmedian. The three symmedians of the triangle ABC concur at the 'Symmedian point'.

4. Eder Contreras Ordenes (or anyone else), Could you please tell me why you are able to state <CAB=<CBP. in your step 2 above? also, is there any way to show that <EDA=<BPD without using that AEDC is cyclic? Thank you very much for any help!

5. You're probably long gone but <CAB = <CBF = <DEB from the alternate segment theorem.