## Monday, May 15, 2023

### Geometry Problem 1539: Unraveling the Lengths in an Isosceles Triangle with Altitude and Tangent Secrets!

Geometry Problem 1539. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

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1. https://photos.app.goo.gl/gmaVeeA2GEGK4kNd8

Locate point F on AC extension such that AC=CF
Note that ABF is a right triangle
Power of point A to circle O = AE.AC= AD^2= 100
Relation in right triangle ABF
AB^2=AE.AF= 2.AE.AC= 200
So AB= 10.sqrt(2)

2. If M is midpoint of AB, CBME is cyclic, thus AM.AB=AE.AC=AD^2, or AB^2=200, AB=10 sqrt (2)

3. Let M be midpint of AB.
BCEM is concyclic.
AM.AB = AE.AC = AD^2
AB^2/2=100 => AB = 10 sqrt 2.

4. A more general version will be:
AC = BC, BE is altitude, AD is tangent to a circle passing through E and C. Show that AB = AD sqrt 2