Question

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 $AC.$

Answer 1

`6.35 \:or\:(11√(3) )/(3)`

Explanation:

Hi there!

1) Well, after reading, checking the measures we can see that this an isosceles triangle. In any Isosceles Triangle, the following line segments do coincide with the perpendicular bisector, the median, altitude.

That is why the perpendicular bisection, the midpoint, coincide. Check picture

2) In addition to this, we can use a formula so that we can find the height of this isosceles triangle based on their equal sizes:

`h=(1)/(2)√(4a^2-b^2)\rightarrow h=(1)/(2)√(4(22)^2-(22)^2) \therefore h=11√(3)\approx19.05`

3) As mentioned, the Median coincides with the altitude in this Isosceles triangle. In the Median, there is a relationship between the segments after the centroid. One line segment is 2/3 bigger of the whole altitude, and the lesser line segment is 1/3 of the height.

We can say that:

Altitude=19.05

`d_{E \:to\:\overline{AC}} <d _(E\:to\:B)\\(1)/(3)*19.05 <(2)/(3) *19.05 \\d_{E \:to\:\overline{AC}}=6.35`

Answer 2

ME = EP = 12

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.