Hello, Rinnie!

The last one is true.

First, we write the convese and see if it is also true,

. . or try to find a counterexample.

If the quadrilateral is an isosceles trapezoid,

. . then a pair of base angles are congruent. Converse:

If a quadrilateral has a pair of equal base angles, it is an isosceles trapezoid.

Code:

*
* *
* *
* *
* *
* *
* *
* θ θ *
* * * * * * * *

This is a quarilateral with equal base angles,

. . but it is **not** an isosceles trapezoid.

The converse is false.

If the quadrilateral is a kite,

. . then the nonvertex angles are congruent. Converse

If the nonvertex angles of a quadrilateral are congruent,

. . the quadrilateral is a kite. Code:

*
* *
* *
* *
* θ *
* *
* θ *
* *
* *
* *
*

This is a quadrilateral with congruent nonvertex angles,

. . but it is **not** a kite.

The converse is false.

If the quadrilateral is a kite,

. . then the major diagonal bisects the vertex angles. Converse:

If the major diagonal of a quadrilateral bisects the vertex angles,

. . then the quadrilateral is a kite. Code:

A
*
*|*
* | *
*α | α*
* | *
* | *
B * | * D
* β | β *
* | *
*
C

There is no way to avoid drawing a kite.

We have: .$\displaystyle \begin{Bmatrix}\angle BAC \:=\: \angle DAC \:=\: \alpha \\ AC \:=\:AC \\\angle BCA \:=\:\angle DCA \:=\:\beta \\ \therefore \Delta ABC \cong \Delta ADC \\ (a.s.a) \end{Bmatrix}$

Therefore, $\displaystyle ABCD$ is a kite.

The converse is true.