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

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

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## Sunday, October 30, 2016

### Geometry Problem 1281 Triangle, Circle, Diameter, Tangent, Altitude, Congruent Angles

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https://goo.gl/photos/cZVha9LddjDaWn519

ReplyDeleteConnect OB, OD ,OE where O is the midpoint of AC

We have OD⊥BD and OE⊥BE and OB is the angle bisector or angle BDE

So O, H, D, B, E are cocyclic

In circle OHDBE ∠AHD= ∠DBO

∠CHE=∠OHE= ∠OBE

But ∠DBO=∠OBE => ∠AHD=∠CHE

Let AD, HB meet at P. Let HPB extended meet AD extended at X.

ReplyDeleteNow P is the orthocenter of triangle AXC, hence AP is perpendicular to XC.

But AE is perpendicular to XC. It follows that A,P,E are collinear and XEC

are collinear.

P being the orthocenter of triangle AXC, easily P is the incentre of

triangle HDE.

So < AHD = < CHE

Sumith Peiris

Moratuwa

Sri Lanka

Problem 1281

ReplyDeleteSuppose that the DC intersect BH in point P and AD the BH to Q.Then < DCA=<QDB ( cord and tangent) and <DQB=<DCA (perpendicular),so <DQB=<QDB or BD=BQ=BE. But <PDB=90-<BDQ=<DPB,then BD=BP=BE=BQ. So PE is perpendicular to BE.

Then at triangle PAC the point P is orthocenter, and A,P and E

Are collinear. The ADPH ,HPEC are concyclic. So <DHA=<DPA=<EPC=<EHC.

4 HIGH SCHOOL OF KORYDALLOS PIRAEUS-GREECE

m(BOD) = mBOE) = m(BOH) = 90 => B,D,H,O & E are concylic

ReplyDeleteSince BD = BE => m(BHD) = m(BHE) = @ (say)

Therefore m(AHD) = m(CHE) = 90-@