In trapezoid ABCD with bases AD and BC, the bisector of angle BAD passes through the midpoint M

In trapezoid ABCD with bases AD and BC, the bisector of angle BAD passes through the midpoint M of side CD. It is known that AB = 5, AM = 4. Find the length of the segment BM.

1. Draw a segment KM parallel to the bases of the trapezoid. Since KM and the bases of AD and BC are parallel, and point M is the midpoint of CD, then KM is the middle line of trapezoid ABCD ⇒:
AK = BK = AB / 2 = 5/2 = 2.5.
2. The bisector AM is a secant intersecting two parallel lines KM and AD ⇒ ∠MAD = ∠KMA as criss-cross.
AM divides ∠BAD into two equal angles, then ∠MAD = ∠BAM (aka ∠KAM) ⇒ in △ AKM ∠KAM = ∠KMA ⇒ △ AKM isosceles: AK = KM = 2.5.
3. KH – height △ AKM, drawn to the base. Since △ AKM is isosceles, then KH is both the height and the median and the bisector ⇒ AH = MH = AM / 2 = 4/2 = 2.
By the Pythagorean theorem from △ AHK:
KH = √ (AK² – AH²) = √ ((2.5) ² – 2²) = √ (6.25 – 4) = √2.25 = 1.5.
4. From △ BAM: KH is the middle line of △ BAM, since it connects the midpoints of sides AB and AM.
The middle line of the triangle cuts off a triangle similar to it from the original triangle ⇒:
AB / AK = BM / KH;
5 / 2.5 = BM / 1.5;
BM = (5 * 1.5) / 2.5 (proportional);
BM = 3.
Answer: BM = 3.