## Friday, December 13, 2013

### Geometry Problem 940: Circle, Tangent, Secant, Chord, Parallel, Perpendicular, 90 Degrees, Midpoint

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to enlarge the problem 940.

#### 4 comments:

1. Angle ABC =angle BFC =angle AHC
So ABHC concylic and since ABCO concylic by tangent property , we have ABHCO concylic
Angle OHC = angle OCA = 90
The second part is by the property of chord, straight forward
QED

2. Notice that BDCE is a harmonic quadrilateral, so the cross ratio of pencil (FE, FD ; FC, FB) = -1. So, if we intersect this pencil with line EA, we will obtain a set of harmonic points, but FB cuts EA at the point of infinity, thus (E, D ; H, P∞) = -1, and it follows that H is the midpoint of DE and it follows that OH ⊥ DE.
Proved.

3. http://img21.imageshack.us/img21/2474/uim3.png

Since FB//ED => Arc(EF)=Arc(BD)
We have ∠ (CHD)= ½(Arc(EF)+ Arc(DC))
= ½(Arc(BD)+ Arc(DC))= 1/2Arc(BC)= ∠ (COA)
So quadrilateral OCAH is cyclic => OH ⊥ HA and H is the midpoint of ED

4. Call <A=a. <DBA+<BAD=<EDF+<FDB, but <EDF=<DBA so <BAD=<FDB=b. <OCH=90-(b+90-a/2)=a/2-b=<OAH which makes HACO cyclic.