#### Question

Given quadrilateral ABCD , AD \parallel BC , BD bisects ∠ADC ,CE bisects ∠BCD , intersecting AB at point E , and intersecting BD at point O

Prove the distance from O to EB is equal to that from O to ED .

Given quadrilateral ABCD , AD \parallel BC , BD bisects ∠ADC ,CE bisects ∠BCD , intersecting AB at point E , and intersecting BD at point O

Prove the distance from O to EB is equal to that from O to ED .

\because AD\parallel BC

The two parallel lines are cut by a transversal BD, then the pair of alternate interior angles \angle ADB and \angle CBD are congruent, that is

\angle ADB =\angle CBD

According to the given condition, BD bisects \angle ADC

\angle ADB = \angle CDB

Therefore

\angle CDB = \angle CBD

\therefore \triangle CBD is an isosceles triangle

Since CE bisects \angle BCD

EC\perp BD

and

\triangle CED and \triangle CBE are congruent

BE = DE

Therefore, the distance from O to EB is equal to that from O to ED