See complete Problem 187 at:

www.gogeometry.com/problem/p187_right_triangle_circle.htm

Right Triangle, Altitude, Incenters, Circles, Angles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Thursday, October 2, 2008

### Elearn Geometry Problem 187

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O is the mid-point of EF, K is a point on BD such that KO⊥EF

ReplyDeleteas mFOK=2mFDK and OF=OD, O is the circumcenter of FDK, and hence F,D,E,K concyclic with center O

as mEKF=2EBF and KE=KF, K is the circumcenter of BFE, and hence mKBF=mBFK

extmBFD=mKBF+45°=mBFK+45°=mBFE, and hence mGFB=mBFD, ⊿BFG≡⊿BFD (ASA)

finally we get BH=BD=BG, and B is the circumcenter of HDG

mHDG=270/2=135°

Solution of problem 187.

ReplyDeleteIn problem 186 it was proved that BG = BD. It can be proved similarly that BH=BD. So, H, D and G belong to the same circle with center B. The central angle GBH is right, so the arc GH that doesn’t passes through D measures 270º. The angle HDG subtends that arc, so ang(HDG) = 135º.