Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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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