In triangle $ABC,$ $M$ is the midpoint of $\overline{AB}.$ Let $D$ be the point on $\overline{BC}$ such that $\overline{AD}$ bisects $\angle BAC,$ and let the perpendicular bisector of $\overline{AB}$ intersect $\overline{AD}$ at $E.$ If $AB = 44$ and $ME = 12,$ then find the distance from $E$ to line $AC.$

ME = EP = 12

Step-by-step explanation:

The parameters given are;

In ΔABC

Point M = Mid point of AB

Point D = Point on BC

AD bisects ∠BAC

Perpendicular bisector of AB = ME intersect AD at E

We note that the line PE representing the shortest distance from E to the line AC at P will be perpendicular to AC (Shortest distance from a point to a line = radius of a circle with center at the point and the line as tangent to the circle)

Therefore;

∠EPA = 90°

In ∠MAE = ∠PAE (angle bisected by AD)

∠AME = ∠EPA = 90° (Angle between perpendicular lines)

In ΔEAM, ∠MAE + ∠AME  + ∠AEM = 180°

Similarly in ΔEAP, ∠PAE  + ∠EPA + ∠AEP = 180°

Since,   ∠MAE + ∠AME = ∠PAE  + ∠EPA, we have;

∠AEM = ∠AEP hence;

ΔEAM ≅ ΔEAP ASA (congruent condition)

Since side AE ≅ AE (reflective property)

Therefore;

For right triangles ΔEAM and ΔEAP to be equivalent, ME = EP = 12.

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