Exercise 3 Page 397

Finally, since AMCD is a parallelogram, by the Parallelogram Opposite Sides Theorem we have that CM ≅ AD. Then, the Transitive Property of Congruence tells us that AD ≅ BC, which implies that the trapezoid is isosceles.

Case 2

This time we will consider a trapezoid where the base angles corresponding to the smaller base are congruent.

Since AB ∥ CD and since AD and BC are transversals, by the Consecutive Interior Angles Theorem we have that both ∠ A and ∠ D, as well as ∠ B and ∠ C, are supplementary. Thus, by applying the Congruent Supplements Theorem, we get that ∠ A ≅ ∠ B.

Notice that we've arrived to the same situation we studied in case 1. Then, we conclude that the trapezoid is isosceles.

Kites

Let's consider a kite ABCD.

If we draw the diagonal AC, it will divide the kite into two triangles which have three corresponding congruent sides.

Then, by the Side-Side-Side (SSS) Congruence Theorem, we have that △ ABC ≅ △ ADC. This implies that ∠ B ≅ ∠ D.

Notice that ∠ A cannot be congruent to ∠ C because this would imply that ABCD would have both pairs of opposite angles congruent, and this would cause ABCD to be a parallelogram due to the Parallelogram Opposite Angles Converse. Since a kite is not a parallelogram, this is a contradiction.

Additionally, since △ ABC ≅ △ ADC, we have that ∠ ACB ≅ ∠ ACD and ∠ CAB ≅ ∠ CAD, which means that AC bisects ∠ A and ∠ C.

Next, let's draw the second diagonal BD. This time, the kite will be divided into four triangles.

By the Side-Angle-Side (SAS) Congruence Theorem, we have that △ PCB ≅ △ PCD, which implies that ∠ BPC ≅ ∠ DPC. Notice that these angles form a linear pair, so they both are right angles.

Thus, we can write another property of kites.