Geometry Problem 13: Cyclic Quadrilateral

Problem 13

Let ABCD be a cyclic quadrilateral. AD and BC, when extended, intersect at a point E. Similarly, AB and CD, when extended, intersect at a point F. Show that:

AE x BF = BE x CF

Here’s a simulation that demonstrates the problem.

Try dragging the vertices of the quadrilateral and observe the products AE x BF and BE x CF. Are they always equal?

Solution

Here’s what you’ll need to know to solve this problem.

Sine Rule

In any triangle ABC,

BC/sinA = CA/sinB = AB/sinC

Here’s a simulation where you can explore this property.

You can drag the three vertices and observe the three ratios BC/sinA, CA/sinB, and AB/sinC. Are they always equal?

Cyclic Quadrilaterals

The sum of the opposite angles of a quadrilateral is always 180°. Here’s a simulation that demonstrates this property.

You can drag the vertices and observe the sum of the opposite angles. Is the sum always 180°?

To proceed, you can apply the sine rule such that the sides AE, BF, BE, and CF are involved. Then, you’ll need to use the above property of cyclic quadrilaterals, to get rid of the terms involving sines of angles.

That’s all for this problem. I suggest you proceed on your own from this point. Good luck!

Were you able to solve it using the above hints? Or, did you solve using any other method? Or, do you need some more help? You can reach out at the Telegram group.

Problem Source: CutTheKnot

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