4. Let the intersection be O CAD=CBD, so AO=BO AC is parallel to BD, so ADB=CAD=CBD, so BO=DO, so AO=CO, so AD=BC.

In the triangle ABC ABD. Both are right triangles. HL can prove congruence. So the topic of AC=BD can be proved.

5. In the same way [4], it can be proved that the triangle ADE ADF coincides with HL [hypotenuse and right angle]. The title can prove

6. If the distances between three sides are equal, let three points be ABC and find the bisector of any two angles.

7. Equality. Because it is the same speed and time, we can get AC=BD, and because it is parallel, the angle CAB= angle ABD is equal. Angle, triangle congruence. Solve.

8. Not much to say. Edge can prove that triangle ABC and triangle DEF are congruent. So the angle ACB= the angle DFE is equal, and the two straight lines are parallel. So AC is parallel to DF and angle ABC= angle DEF, so straight line AB is parallel to DE.