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
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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