Geometry Problem 1539. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Get the full picture! Click below to uncover the details.

Click for additional details.

Share your solution by clicking 'Comment' below the post or entering your solution or comment in the 'Enter Comment' field and pressing 'Publish'.

Share your solution by clicking 'Comment' below the post or entering your solution or comment in the 'Enter Comment' field and pressing 'Publish'.

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

ReplyDeleteLocate 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)

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

ReplyDeleteLet M be midpint of AB.

ReplyDeleteBCEM is concyclic.

AM.AB = AE.AC = AD^2

AB^2/2=100 => AB = 10 sqrt 2.

A more general version will be:

ReplyDeleteAC = BC, BE is altitude, AD is tangent to a circle passing through E and C. Show that AB = AD sqrt 2