Geometry Problem

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

This entry contributed by Ajit Athle.

Click the figure below to see the problem 767 details.

## Monday, June 18, 2012

### Problem 767: Equilateral Triangle, Circle tangent to a side, Tangent line, Measurement, Secant

Labels:
circle,
equilateral,
measurement,
secant,
tangent,
triangle

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By theTangent Theorem:

ReplyDeletem^2 = (m+n-d)(m+n-e); n^2=(m+n-g)(m+n-f)

By the Secant Theorem:

de = fg or fg - de = 0

Then

m^2 - n^2 = (m+n)(f+g-d-e)

(m+n)(m-n) = (m+n)(f+g-d-e)

m-n = f+g-d-e

So:

m + d + e = f + g + n

http://www.mathematica.gr/forum/viewtopic.php?f=22&t=27794&p=135568

ReplyDeletehttps://goo.gl/photos/FUFbuawj6LN7feiG6

ReplyDeleteLet N and P are the projection of O over AB and AC

Let M is the midpoint of AC

Note that N and P are the midpoints of ED and FG

Observe that MT= ½(m-n)

BN=1/2(d+e)=BQ

BP=1/2(f+g)

PQ=BP-BQ= ½(f+g-d-e)

Since ABC is equilateral so angle(AC, NO)=angle (AO,BC)= 30 degrees

And MT=PQ => ½(m-n)= ½(f+g-d-e)

So m+d+e=n+f+g