Let AB and CD be the given lines with intersection E.
We will show that angle AEC equals angle DEB.
Since AE stands on CD, it follows that angles
CEA and AED are two right angles.
(Prop I-13)
Similarly, since DE stands on AB, the angles BED
and DEA are two right angles.
(Prop I-13)
Thus angles BDE and DEA are equal to angles CEA
and AED.
(Post-4 and CN-1)
Angle BDE is therefore two right angles minus angle DEA. Also angle CEA is two right angles minus angle AED.
Since equals subtracted by equals are equal (CN-3), it follows that angle CEA equals angle BED, which was to be proved.
QED
QED
![]() |