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 $$ ED $$ .

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Uniteasy Admin · Accepted Answer

$$\because AD\parallel BC $$
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
$$\angle CDB = \angle CBD$$
$$	herefore 	riangle CBD$$ is an isosceles triangle
Since $$CE$$ bisects $$\angle BCD$$
$$EC\perp BD $$
and
$$	riangle CED$$ and $$	riangle CBE$$ are congruent 
$$BE = DE$$
Therefore, the distance from $$ O $$ to $$ EB $$ is equal to that from $$ O $$ to $$ ED $$

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