Geometry Problem

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

Click the figure below to see the complete problem 637.

## Tuesday, July 19, 2011

### Problem 637: Semicircle, Diameter, Perpendicular, Inscribed Circle, Chord, Tangent, Arbelos

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AG^2=AD^2 - DG^2

ReplyDelete=(2R-r)^2 - r^2

=4R^2 - 4Rr

=4R(R-r)

AE^2=AC.AB (tr AEC and tr ABE are similar)

=(2R-2r).(2R)

=4R(R-r)

AE^2=AG^2,

AE=AG

Observe that AEB is a right triangle ( AB is a diameter)

ReplyDeleteso AE^2= AC.AB (1) ( Relation in a right triangle)

In circle center D we also have

AG^2= AC.AB (2) ( power of point A to circle D)

Compare (1) and (2) we get AE=AG

Peter Tran

Using my proof and the result of Problem 636 AG is a tangent to circle CGB, hence AG^2 = AC.AB

ReplyDeleteBut AE^2 also = AC.AB since AE is a tangent to circle BCE

So AE = AG

Sumith Peiris

Moratuwa

Sri Lanka