$18 \cdot 22$ As shown in the figure, in quadrilateral $A B C D$, $\angle A=\angle C$. Then the quadrilateral $K L G H$ formed by the intersection of the angle bisectors of each interior angle is
(A) A quadrilateral with only two equal angles.
(B) A parallelogram.
(C) A trapezoid.
(D) An isosceles trapezoid.
(Anhui Province, Anqing and other counties and cities Junior High School Mathematics Competition, 1991)
[Solution] Given $\angle A=\angle C$, then $\angle D A G=\angle D C H$.
$$
\begin{array}{ll}
\text { Also } & \angle A D G=\angle C D H, \\
\therefore & \triangle A D G \backsim \triangle C D H . \quad \therefore \quad \angle D H C=\angle A G D . \\
\text { Also } & \angle D H C=\angle K H G, \quad \therefore \quad \angle K H G=\angle L G H . \\
\text { By } & \angle B K C+\angle A G D \\
& =\frac{1}{2}(\angle A+\angle B+\angle C+\angle D) \\
& =180^{\circ}, \\
\therefore & \angle L K H+\angle K H G=180^{\circ}
\end{array}
$$

Therefore,
$K L \parallel G H$.

In summary, quadrilateral $K L G H$ is an isosceles trapezoid. Hence, the answer is $(D)$.