See complete Geometry Problem 134

Orthic Triangle, Altitudes, Angle Bisectors, Orthocenter, Incenter. Level: High School, SAT Prep, College geometry

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## Monday, July 21, 2008

### Elearn Geometry Problem 134

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Quad AFHE is cyclic since angle AFH=90=angle AEH

ReplyDeleteSimilarly quad AFDC is cyclic since arc AC subtends equal angles of 90 deg. at F & D. Further quad EHDC is also cyclic since opposite angles are each=90. From cyclic quad EHDC we've angle HDE = angle HCE. From cyclic quad AFDC we've angle FDA (or FDH) = angle FCA (or HCE) Hence angle FDH = angle HDE. In other words HD (or AD) bisects angle FDE -----(2). It can be similarly proven that HF bisects angle EFD and HE bisects FED which makes H the incenter of triangle DEF.-----(3)

Now quad AFDC being cyclic angle A = angle BDF while angle ADB=90=angle ADF + angle BDF=angle ADE + angle EDC. As proven earlier angle FDA = angle ADE which makes angle BDF = angle EDC both being equal to angle A.------(1)

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Solution of problem 134.

ReplyDeleteLet’s write a for alpha and q for theta.

(1) We have ang(ABE) = 90 – a, so ang(BHF) = ang(CHE) = a.

Quadrilateral BDHF is cyclic, since two opposite angles are right angles.

Then a’ = ang(BDF) = ang(BHF) = a.

In the same way, quadrilateral CEHD is cyclic, so a” = ang(EDC) = ang(CHE) = a.

Thus a’ = a” = a and ang(BDF) = ang(EDC).

(2) As q + a’ = q + a = 90 and q’ + a” = q’ + a =90 then q = q’ and AD is the angle bisector of EDF.

(3) Similarly we can prove that BE and CF are the angle bisectors of DEF and DFE, respectively. Then H is the incenter of triangle DEF.

2- Is JUST Blanchet's theorem.

ReplyDelete3- apply (2) for all <F, <D and <E, then H is the incenter of trDEF.