## Saturday, December 24, 2016

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

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

Details: Click on the figure below. 1. Draw TO. Ang HOI=2A, ang ITP=45-A => ang HIO = 90 - 2A
=> IHO = 90

2. https://goo.gl/photos/yYwMDZZwhsdpYQ2w9

Let 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

3. Problem 1298
1. 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

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