<Qua-2025-781>, hint
<Tri-2025-2249>, hint
Let M be the center of arc AB, opposite of C.
Since CM//NQ arc CQ=arc MN, <BCN=<CNQ=<CBQ=15, and <CQB=180- <CH(2)B
<CH(2)B=60. Triangles CQB, BNA, APC are congruent.
Since <MAC=90 NP//MA, arc MN=arcAP=arcNB, triangles NBQ and PAN are congruent
and since arc AMB=arc PAM (arc PA=arcNM) . Smilarly arcPAM=arcCQN
Triangle NQP is equilateral. <AH(1)B=<BH(2)C=<CH(3)B =180-<ANB=180-<BQC=180<-CPA
=60 . H(1),B,H(2) are collinear because <H(2)BQ=90-60=30, <CBQ=15, <ABC=60, <ABH(1)=90--15(=<ABN)=75. Finally triangle H(1)H(2)H(3) is equilateral
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