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

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

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## Saturday, December 24, 2016

### Geometry Problem 1298 Arbelos, Semicircles, Diameters, Circle, Incircle, Tangent, Angle Bisector, Perpendicular, Midpoint

Labels:
angle bisector,
arbelos,
circle,
diameter,
incircle,
perpendicular,
semicircle,
tangent

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Draw TO. Ang HOI=2A, ang ITP=45-A => ang HIO = 90 - 2A

ReplyDelete=> IHO = 90

https://goo.gl/photos/yYwMDZZwhsdpYQ2w9

ReplyDeleteLet D is the midpoint of arc AC with ∠AOD= 90 ( see sketch)

Since TP is the angle bisector of angle ATC => TP will pass thru D

Since Triangles TIP and TOD are isoceless and angle ITP= angle OTD=> Angle TOD= angle TIP

So IP//OD and IP ⊥AC

Problem 1298

ReplyDelete1. Let K medium the arc AB and L medium the arc BC.Is <BTC=<BTA=45 then the points B,P ant T are collinear.But <TAC+<TCA=90 or (45-<KAT)+(<ACT-45)=90 or <KAT=<ACT=<LTC=<KTA so <KTL=90. Is <KTO=<KTA+<ATO=<KAT+<TAO=45=<OTL.So <KTA=<ITP=<IPT=<BIH. But <BIH+<IBH=<KAT+45+<TAB=45+45=90. So IP is perpendicular in AC.

APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

I do not see anybody have solution 0f 2nd part of the question " show that P is the midpoint of IH "

ReplyDelete