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

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

Details: Click on the figure below.

## Monday, July 16, 2018

### Geometry Problem 1365: Circle, Tangent, Secant, Chord, Parallel, Midpoint, Sketch, iPad Apps

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< AMB = < BFC < ACB = < AOB since ABOC is concyclic

ReplyDeleteHence ABMO is concyclic and so < AMO = < ABO = 90

Therefore M is the midpoint of DE

Sumith Peiris

Moratuwa

Sri Lanka

https://photos.app.goo.gl/BGn1ZyJuoHQCZUsQ7

ReplyDeleteConnect OA, OB, OC

Since AB and AC tangent to circle O => ∠ (BOA)= ∠ (COA)

We have ∠ (BFC)= ½. ∠ (BOC)= ∠ (BOA)

∠ (BFC)= ∠ (BMA)= ∠ (BOA) => MOAB is cyclic quad.

So ∠ (OMA)= ∠ (OBA) = 90 degrees => M is the midpoint of ED