An Olympiad Problem!

Geometry Level 5

Trapezoid $ABCD$ has right angles at $C$ and $D$ , and $AD$ > $BC$ . Let $E$ and $F$ be the points on $AD$ and $AB$ , respectively, such that $\angle BED$ and $\angle DFA$ are right angles. Let $G$ be the point of intersection of segments $BE$ and $DF$ . If $\angle CED = 58^{\circ}$ and $\angle FDE = 41^{\circ}$ , what is $m\angle GAB$ ?

Disclaimer: This was an olympiad problem. All credits to the one who originally made this problem.

The answer is 17.

2 solutions

Manuel Kahayon
Apr 7, 2016

We start off by drawing $BD$ . Then we extend $AG$ to a point $H$ on $BD$ . Since $DF$ and $BE$ are both altitudes of $\triangle ABD$ , then $G$ is the orthocenter. Similarly, we conclude that $AH \perp BD$ . That means $AFHD$ is a cyclic quadrilateral and $m\angle GAB = m\angle FDB = 58 - 41 = 17.$

Same problem has appeared a few months back.

Niranjan Khanderia - 5 years, 2 months ago

How have you concluded AFHD is cyclic ?? Please explain.

Chirayu Bhardwaj - 5 years, 2 months ago

From the fact that $\angle DFA = \angle AHD = 90^{\circ}$ , you can immediately conclude that by Thales' theorem , $AFHD$ is cyclic.

Reineir Duran - 5 years, 2 months ago

Ahmad Saad
Apr 7, 2016

Basically, I did the same way. But I think it should be shown how quadrilaterals are cyclic, specially BDEF.

Niranjan Khanderia - 5 years, 2 months ago

One of necessary and sufficient condition for a convex quadrilateral to be cyclic is that

an angle between a side and a diagonal is equal to the angle between the opposite side and the other diagonal.

That is, for covex quadrilateral BDEF :

angle between side DE and diagonal EB = <DEB = 90 degrees.

angle between opposite side BF and the another diagonal FD = <BFD = 90 degrees.

sinse, the two angles are equal which satisfy that codition, then the quadrilateral BDEF is cyclic.

Ahmad Saad - 5 years, 2 months ago

Thank you. I did not know this. So it was a long way I did.

Niranjan Khanderia - 5 years, 2 months ago

An Olympiad Problem!

The answer is 17.

2 solutions

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