Read full issue - Canadian Mathematical Society
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270<br />
E<br />
F<br />
A<br />
B H C G<br />
Now BF = GF and \F BG = \FGB =40, so that \IGF =20.<br />
Also \IF G = \FBG+\FGB =80 , so that<br />
\F IG = 180 , \IF G , \IGF<br />
D<br />
= 180 , 80 , 20<br />
= 80 :<br />
Therefore 4GIF is an isosceles triangle, so<br />
I<br />
GI = GF = BF : (1)<br />
But 4BGI and 4ABC are congruent, since BG = AB,<br />
\GBI = \BAC; \BGI = \ABC.<br />
Therefore<br />
From (1) and (2) we get<br />
So in 4BCF,<br />
\BCF =<br />
GI = BC : (2)<br />
BC = BF :<br />
180 , \F BC<br />
2<br />
= 180 , 40<br />
2<br />
= 70 :<br />
Thus \F CB =70 and that proves that the given lines CE and BD<br />
and the perpendicular line AH meet at one point.<br />
Solution 2 by Adrian Birka, Lakeshore Catholic High School, Port<br />
Colborne, Ontario.<br />
First we prove the following lemma:<br />
In 4ABC; AA 0 ; BB 0 ; CC 0 intersect if and only if
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