**The problem:**
Let A be an outside point from the circle with the center O. Draw the secant ABC of the mentioned circle. The tangents at B and C got an intersection, named K. From K, we draw the perpendicular to AO. Name the intersection H. E and F are 2 intersections of KH and the circle with the center O. (E is between K and F). We name M for the intersection of KO and BC. Prove that:

- There exists a circle that pass throught 4 points E, M, O and F
- AE and AF are tangents of the circle with center O

**Notes:**

- I tried to translate this problem from my language to English. So If I use wrong grammar structure or wrong expression, please forgive me.
- This is not my homework. I like to look for and solve "hard" problems, by my view. This is one of those, and I can't find out the way to solve it.

## Comments

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TopNewestThis problem can be solved by the similarity relationships in a right triangle with the altitude of the hypotenuse. Here's a sketch of my solution: 1. By the power of a point theorem(or whatever you want to call it, it's basically simiarity), we get that KExKF=KB^2. Moreover, since ∠KBO=90° and BM⊥KO, we get KMxKO=KB^2=KExKF→O,M,E,F are concyclic. 2. We know that AE is tangent to circle O iff ∠OEA=90°. Since ∠KMA=∠KHA=90°, so K,M,H,A are concyclic. Thus OHxOA=OMxMK=OB^2=OE^2→By SAS similarity △EOH is similar to △AOE→∠OEA=∠EHO=90° and that proves AE is tangent to circle O. By analogy AF is also tangent to circle O. – Xuming Liang · 3 years, 12 months ago

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– Xuming Liang · 3 years, 12 months ago

sorry for not following the formatting guidline :(Log in to reply

– Đức Việt Lê · 3 years, 12 months ago

Thanks a lot! I can understand the solution that you gave me. But I cannot figure it out :DLog in to reply