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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Details: Click on the figure below.
Draw TO. Ang HOI=2A, ang ITP=45-A => ang HIO = 90 - 2A=> IHO = 90
https://goo.gl/photos/yYwMDZZwhsdpYQ2w9Let 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 DSince Triangles TIP and TOD are isoceless and angle ITP= angle OTD=> Angle TOD= angle TIP So IP//OD and IP ⊥AC
Problem 12981. 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