Geometry prob

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.

Easy Math Editor

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Comments

This 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 - 8 years ago

sorry for not following the formatting guidline :(

Xuming Liang - 8 years ago

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

Đức Việt Lê - 8 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$