# 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:

1. There exists a circle that pass throught 4 points E, M, O and F
2. 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.

Note by Đức Việt Lê
5 years, 4 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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.

- 5 years, 4 months ago

sorry for not following the formatting guidline :(

- 5 years, 4 months ago

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

- 5 years, 4 months ago