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Based on the figure, the following conditional statement holds true.

$⎩⎪⎪⎪⎨⎪⎪⎪⎧ ADbisects∠BACDB⊥ABDC⊥AC ⇒DB=DC $

Consider an angle and its bisector.

Let $D$ be a point lying on the bisector of the angle. Also, let $DB$ and $DC$ be the distances from $D$ to the sides of the angle. Recall that the distance from a point to a line is perpendicular to the line.

Since $AD$ bisects $∠BAC,$ by the definition of an angle bisector it can be said that $∠BAD$ and $∠CAD$ are congruent angles. Furthermore, $∠ABD$ and $∠ACD$ are both right angles. Therefore, they are also congruent angles.$∠BAD≅∠CADand∠ABD≅∠ACD $

By the Reflexive Property of Congruence, $AD$ is congruent to itself.
Because two angles and a non-included side of $△ADB$ are congruent to two angles and the corresponding non-included side of $△ADC,$ the triangles are congruent by the Angle-Angle-Side Congruence Theorem.
$△ADB≅△ADC $

Corresponding parts of congruent triangles are congruent. Therefore, $DB$ is congruent to $DC.$ Congruent segments have equal measures.
$DB=DC $

This means that $D$ is equidistant from the rays of $∠BAC.$