## Tuesday, April 22, 2014

### Geometry Problem 1006: Tangent Circles, Common External Tangent, Bicentric Trapezoid

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

Click the figure below to see the complete problem 1006.

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4. Problem 1006 has been replaced. Thanks.

5. Since EA = ET = EB, FC = FT = FD and AB = CD.
Also AD//BC so ABCD is an isosceles trapezoid.
Then EF = (AD + BC)/2. (see the solution of problem 1008).
EF = AB = CD so AB + CD = AD + BC then ABCD is circumscribed quadrilateral .
Any isosceles trapezoid is a cyclic quadrilateral.

6. Fie E,F ,G si H mijloacele segmntelor AB,BC,CDsi respectiv AD=>EFGH patrat,(deoarece ABCD este un trapez isoscel si are diagonalele congruente)

7. Extend AB, DC to meet at E. Then EBA and ECD are tangents drawn to the 2 circles from an external point E and are hence equal for each circle.

So Tr.s EAD & EBC are isoceles and similar since they share the same angle <E. So BC//AD and < B = < D. Hence ABCD is a cyclic trapezoid with AB = CD.

The circumcentre of ABCD must lie on OQ the perpendicular bisector of both AD and BC. Find O1 on OQ such that AO1 = BO1 and we can see easily thro congruence that these are also equal to DO1 and CO1. Hence O1 is the circumcentre

Now draw the common internal tangent at T which is seen to bisect AB and CD so AT is perpendicular to BT and likewise DT to CT. Hence thro congruence we can show that the perpendiculars from T to ths 4 sides are all equal hence ABCD has an incircle with C as centre

Sumith Peiris
Moratuwa
Sri Lanka

8. Sorry typo.

Last line should be "with T as centre"