Question

Given: Be bisects <ABC CE bisects <DCB, and m<1 +m<2 =90
Prove BA II CD

Answer 1

Solution.

We have $\Delta ABC$ . We know that $BE$ bisects $\angle ABC$ . Then $\angle EBC = 45{}^{\circ}$ (as a bisector) and $\angle BEC = 90{}^{\circ}$ (as a height). So $\angle BCE = 180{}^{\circ} - 45{}^{\circ} - 90{}^{\circ} = 45{}^{\circ}$ . We know that $CE$ bisects $\angle DCB$ and from this it follows that $\angle ECD = 45{}^{\circ}$ . So we have a rectangular $ABCD$ . It has two sides $AB, CD$ which are parallel.

So $BA||CD$