Advanced Geometric Problem & Solution (Twice a week update) Now 4641 problems posted.
<Tri-2025-2193>, hint
Since HC is diameter of the circumcircle of HB'CA', A'B'=HC sin45
angles ACB= CAA'=CBB'=45
CH^2=A'C^2+HA'^2(=A'B^2).
Since A'E=A'D=A'B' DE^2=2A'B'^2 (A'B'=HC sin45). Hence CH=DE
No comments:
Post a Comment