O is the mid-point of EF, K is a point on BD such that KO⊥EF as 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. In 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.
From problem 186, BD=BG and with the same rationale behind, BH=BD <BDG=<BGF=x <DBC=180-2x <DBA=2x-90 <BDH=(180-(2x-90))/2=135-x <HDG=<BDH+<BDG=135-x+x=135
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.
From Problem 186, B is the circumcentre of Tr. HDG and since < B = 90, the result follows
ReplyDeleteSumith Peiris
Moratuwa
Sri Lanka
From problem 186, BD=BG and with the same rationale behind, BH=BD
ReplyDelete<BDG=<BGF=x
<DBC=180-2x
<DBA=2x-90
<BDH=(180-(2x-90))/2=135-x
<HDG=<BDH+<BDG=135-x+x=135