Geometry Problem. Post your solution in the comments box below.

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

Click the diagram below to enlarge it.

## Wednesday, July 15, 2015

### Geometry Problem 1137: Triangle, Circumcircle, Orthocenter, Midpoint, Arc, Collinear Points, Tangent Circles

Labels:
arc,
circle,
circumcenter,
circumcircle,
collinear,
diameter,
midpoint,
orthocenter,
tangent,
triangle

Subscribe to:
Post Comments (Atom)

http://s14.postimg.org/j7s4od1u9/pro_1137.png

ReplyDeleteDraw points P, Q, S per attached sketch

Observe that H, M, P are collinear and AC is the perpendicular bisector of HQ. ( see other problem)

With manipulation of angles we will get ∠ (EHF)= ∠ (BPF)= ∠ (HQS)= ∠ (QHS)= ∠ (SMH) …( see sketch)

We will get the following results:

HF//AC and MH tangent to circumcircle of triangle QHF

S is the circumcenter of triangle QHF

In right triangle MHS with altitude HD we have SH^2=SF^2=SD.SM

So SF tangent to circuncircle of MDF

SH ⊥ HE and SH=SF => SF tangent to circumcircle of HFE

So SF is tangent to both circles O1 and O2