<ABC=2<BAC wanted, BD// EF, angle bisectors (2015 Euler Olympiad Remote 2.2)

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<ABC=2<BAC wanted, BD// EF, angle bisectors (2015 Euler Olympiad Remote 2.2)

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#1 h Feb 10, 2021, 12:30 PM • 3 Y

Y by Mango247, Mango247, Mango247

In the triangle $ABC$ the angle bisector $BD$ was drawn, in the triangle $BDC$ the angle bisector $DE$ , and in the triangle $DEC$ the angle bisector $EF$ . It turned out that the lines $BD$ and $EF$ are parallel. Prove that the angle $ABC$ is twice the angle $BAC$ .

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#2 h Feb 10, 2021, 12:58 PM

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Let angle $DBC$ be $x$ and angle $BDE$ be $y$ .Applying angle sum property in $\triangle DEB$ ,we get angle $DEB$ is equal to $180-(x+y)$ .Also,angle $DEF$ is equal to $x$ and angle $DFE$ is equal to $180-(x+y)$ and by linear pair,angle $EFC$ is equal to $x+y$ .But,from properties of parellel lines,we know that angle $EFC=2y$ .Equating the two expressions,we get $x=y$ .From our initial notation,we get angle $ABC=2x$ .By angle sum property in $\triangle DAB$ ,we get angle $BAC=2y-x=2x-x=x$ .Therefore,angle $ABC=2$ angle $BAC$ .This is what we needed to prove

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