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

Labels:
90,
angle,
circumcircle,
degree,
midpoint,
orthocenter,
triangle

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http://s17.postimg.org/3wo281qsv/pro_1136.png

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