See complete Problem 37

Right triangle, altitude, incenters, orthocenter. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 37

Labels:
altitude,
incenter,
orthocenter,
right triangle

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join A → F → E, C → G → E

ReplyDeleteextend FE → E' (on BG), GE → E" (on FB)

ang ABD = C, ang DBC = A ( perpendicular sides )

▲CE"G =>

ang E"CB + ang E"BC = C/2 + A + C/2 = C + A = 90°

=> CE"B = 90°

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▲'AE'B =>

ang E'AB + ang E'BA = A/2 + C + A/2 = A + C = 90°

=> AE'B = 90°

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In problem 25, BE and FG are congruent and perpendicular.

ReplyDeleteLet FP be an altitude on the base BG.

Then we get

(1) ang(A) +ang(C)=90 degree

(2) ang(GFP)=ang(A)/2 +ang(C)/2=45 deg

(3) ang(FGP)=ang(C)/2 + (90- ang(C))/2=45 deg.

Hence we see that ang(EPG)=90 degree and so

E is an orthocenter of triangle BFG.

Extend BF &BG to M and N on AC ( See sketch)

ReplyDeleteNote that AE, CE , BF and BG are angle bisectors , A,F,E and E,G, C are collinear

∠ (M)= ∠ (MBC) …. Both angles complement to ∠ (MBD)

∆ MBC is isosceles => angle bisector CE ⊥ to the base BM

Similarly AE ⊥ to BM => E is orthocenter of ∆BFG

See below for the sketch

ReplyDeletehttp://img98.imageshack.us/img98/9435/problem37.png