Day 21: This Diagram Looks Familiar

Geometry Level 2

A cyclic quadrilateral $ABCD$ has its sides extended and they do meet, creating points $E$ and $F$ in the diagram.

As shown, $B, D, E, F$ are also concyclic.

Find the angle between line segments $EF$ and $AC$ .

This problem is part of the Advent Calendar 2015 .

The answer is 90.

1 solution

Michael Ng
Dec 20, 2015

$\angle EBF =\angle EDF$ (angles in the same segment)

So $\angle ABC =\angle ADC$ . But $\angle ABC+\angle ADC=180^{\circ}$ so both are right angles.

Therefore in triangle $AEF$ , $C$ is the orthocentre; therefore $AC$ is an altitude and so the angle required is $\boxed{90^{\circ}}$ as required.

EBF=EDF then how ABC = ADC?

Joy Patel - 5 years, 5 months ago

ABC=180 - EBF and ADC=180- EDC as EBF=EDC therfore ABC and ADC are equal

Shreyansh Choudhary - 5 years, 5 months ago

Nice observation about the orthocenter to make the final conclusion.

Calvin Lin Staff - 5 years, 5 months ago

Thank you! I love it when a point turns out to be a well known triangle centre :)

Michael Ng - 5 years, 5 months ago

Too good 👍🙌😀

Biswajit Barik - 4 years, 5 months ago

since EF is diameter for bigger circle then edf=ebf=90 , as a result abc=adc=90 hence both ac and ef are diameter of their respective circle hence required angle is 90

Deep Rawat - 1 year ago

EF doesn't need to be the diameter.

Even if AC and EF are diameters, why does that make the angle between AC and EF to be 90 degrees?

Calvin Lin Staff - 1 year ago

Easier way is to think that both sides are symmetric. So it ought to be $90°$ .

Pranjal Jain - 5 years, 5 months ago

Nice! Symmetry is such a powerful technique when used correctly :)

Michael Ng - 5 years, 5 months ago

That's a bad explanation a proof problem (and for guessing, it doesn't build intuition in this case).

In particular, it is interesting to ask "If all that we know is AC is the diameter of the circle. Why must AC intersect EF at right angles (even if B, D are not symmetric about AC)?"

Calvin Lin Staff - 5 years, 5 months ago

Day 21: This Diagram Looks Familiar

This problem is part of the Advent Calendar 2015 .

The answer is 90.

1 solution

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