Cyclic Quadrilateral

Geometry Level 3

Rectangle $ABCD$ is inscribed in a circle. Point $E$ is on side $DA$ so that $DE=18$ and $EA=6$ . If $CE$ is extended, it meets the circumcircle of the rectangle at point $F$ . Given that $DC=18$ , find $DF$ . If the answer is of the form $a\sqrt{2}$ where $a$ is a positive integer, give your answer as $a$ only.

The answer is 15.

2 solutions

A Former Brilliant Member
May 2, 2017

Relevant wiki: Cyclic Quadrilaterals

Aah so ptolemy is the key. I got stuck after finding the easy calculated sides.

Peter van der Linden - 4 years, 1 month ago

Mrunang Kothari
Jul 28, 2019

First things first, Make lines DF and AF.

Now, there is something that can be observed: One can use Ptolemy's Th. for Quad AFDC if u join A and C. , But I wanted some alternate solution.

See that angles DCE and DEC are 45 deg, because DE=DC=18 and angle CDE is right angle. Also, angle FEA is also 45 deg, because it is the vertically opposite angle of angle DEC.

Now, consider chord, DF, we know that the angles subtended buy the chord on the major arc are equal in measure. So we see that angles DCF and DAF are equal in measure and that too 45 deg each, because angel DCF is 45 deg.

Now in triangle, FAE, EF=FA, because angles FEA and FAE are equal in measure. So, we now can deduce that angle EFA is 90 deg. Now, if AE=6, After applying Pythagoras Theorem, we see that EF=FA=6/sqrt(2).

Final step, consider Triangle ADF, apply cosine rule for angle DAF. After that we see that, DF=15*sqrt(2).

Cyclic Quadrilateral

The answer is 15.

2 solutions

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