Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Wednesday, July 15, 2015
Geometry Problem 1136: Triangle, Circumcircle, Orthocenter, Midpoint, Arc, 90 Degrees, Angle
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ReplyDeleteDraw diameter BE ; connect AE and EC
EC⊥BC => EC//AH
EA⊥BA => EA//CH
So AHCE is a parallelogram => H, M, E are collinear
So BD⊥ MD
Extend HM to X such that HM = MX, so AHCX is a parallelogram
ReplyDeleteHence < AHC = A + C = < AXC, hence X must lie on circle ABC {B + (A+C) = 180}
Therefore BX is a a diameter passing thro O and so < BDH = 90
Sumith Peiris
Moratuwa
Sri Lanka