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BdMO Online Forum • View topic - CGMO 2007/5

CGMO 2007/5

For discussing Olympiad level Geometry Problems
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CGMO 2007/5

Post Number:#1  Unread postby Ananya Promi » Sat Jun 03, 2017 3:21 pm

Point $D$ lies inside triangle $ABC$ such that $$\angle{DAC}=\angle{DCA}=30^o$$ and $$\angle{DBA}=60^o$$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF=2FC$. Prove that $DE$ I $EF$.
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Re: CGMO 2007/5

Post Number:#2  Unread postby Atonu Roy Chowdhury » Sat Jun 03, 2017 8:54 pm

$M$ is the midpoint of $AC$ and $X$ is the midpoint of $AF$. Proving $\angle DEF = 90$ is equivalent to $MDEF$ cyclic or $\angle MEF = \angle MDF = 30$

It is easy to show $\triangle DFX$ is equilateral. So, $ABDX$ is cyclic.
$\angle EFC = \angle BXC = 60+\angle BAD$, so $\angle FEC = 90 - \angle BAD - \angle BCD$
$\angle BEM = \angle EMC + \angle FCE = 60 + \angle BAD + \angle BCD$
So, $ \angle MEF = 180 - \angle FEC - \angle BEM = 30$

Q.E.D.
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