Here’s a really neat problem that demonstrates the usage of lemmas in a geometry proof:

Problem : Show that a trapezoid is cyclic if and only if it is isosceles.

Solution:

Lemma 1: A trapezoid is cyclic if it is isosceles.

Proof of Lemma 1: Label the trapezoid as ABCD. Let BC be the longer base of the trapezoid. Since the trapezoid is isosceles, \angle B = \angle C. Because AD \parallel BC, \angle A + \angle B = 180. Therefore, \angle A = 180^{\circ} - \angle B.

Similarly, \angle D = 180-\angle C.

Since \angle B = \angle C, we must have \angle A = \angle D.

We add \angle A and \angle C to get the result: \angle A + \angle C = 180^{\circ} - \angle B + \angle C = 180^{\circ}.

Since the opposite angles of the trapezoid add up to 180^{\circ}, the isosceles trapezoid is cyclic.


Lemma 2: A trapezoid is iscosceles if it is cyclic.

Proof of Lemma 2: Draw trapezoid ABCD, having AB be the shorter base and CD be the longer base. Draw the circumcircle \omega of the trapezoid.

We quickly see that \angle ADB = \angle ACB since they are both inscribed in the same arc.

We also find that \angle BAC = \angle BDC and \angle ACD = \angle ABD since these two pairs are also inscribed in their respective arcs.

Let \angle BAC = \angle BDC = x and \angle ABD = \angle ACD = y. Since AB \parallel CD, \angle BDC = \angle ABD. Therefore, x=y.

So \angle D = x+y = \angle C, which proves that ABCD is an isosceles trapezoid.


Combining Lemma 1 and Lemma 2 gives us the desired result. \blacksquare.

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