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

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

Details: Click on the figure below.

## Friday, February 22, 2019

### Geometry Problem 1412: Right Triangle, Incircle, Excircle, Tangency Points, 45 Degree Angle

Labels:
45 degrees,
angle,
excircle,
geometry problem,
incircle,
right triangle,
tangency point

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Isosceles trapezoid (see P 1411) is cyclic

ReplyDeletehttps://photos.app.goo.gl/AkZzfxhqUxSaGJcL7

ReplyDeleteSee sketch for position of points K, L, M

We have AK⊥DG and

∠ (AIC)= 90+ ½* ∠ (B)= 135 => ∠ (EIC)=45

Since ID and CF perpendicular to AB => DI//FC

We also have CF=MB=DI => DICF is a parallelogram

So DF//IC => ∠ (DLK)= ∠ (EIC)= 45 => ∠(GDF)= 45

If AC meets the excircle at Z, from my Proof of Problem 1407, DFZ are collinear

ReplyDeleteSo < GDA = < AGD - < GZD

= (90 - A/2) - C/2 = 45

Sumith Peiris

Moratuwa

Sri Lanka