In a rectangular trapezoid ABCD with bases AD = 12, BC = 8 and BAD = 90 °, a large diagonal BD = 13.

In a rectangular trapezoid ABCD with bases AD = 12, BC = 8 and BAD = 90 °, a large diagonal BD = 13. Diagonals intersect at point M. a) Prove that triangles BMC and DMA are similar; b) Find the perimeter of triangle ABM.

Let us prove the similarity of the triangles BMC and DMA. In triangles, the angles BMC and DMA are equal as vertical at the intersection of straight lines AC and ВD. Angle BCM = MAP as criss-crossing angles at the intersection of parallel lines BC and BP of the secant AC. Then the triangles BMC and DMA are similar in two angles, which was required to be proved.

From the right-angled triangle ABD we determine the length of the leg AB. AB ^ 2 = BD ^ 2 – AD ^ 2 = 169 – 144 = 25. AB = 5 cm.

From the right-angled triangle ABC, we determine the length of the hypotenuse AC. AC ^ 2 = BC ^ 2 + AB ^ 2 = 64 + 25 = 89. AC = √89 cm.

Let the length of the segment BM = X cm, then DM = (13 – X) cm.Then BC / AD = BM / DM.

8/12 = X / (13 – X).

12 * X = 156 – 8 * X.

20 * X = 104.

X = BM = 104/20 = 5.2 cm.

Let the length of the segment CM = X cm, then the length of the segment AM = AC – X = √89 – X.

Then BC / AD = CM / AM.

8/12 = X / (√89 – X).

12 * X = 8 * √89 – 8 * X.

20 * X = 8 * √89.

X = 8 * √89 / 20 = 2 * √89 / 5.

AM = √89 – 2 * √89 / 5 = √89 * (1 – 2/5) = 5.66 cm.

Determine the perimeter of the triangle ABM.

Ravm = AB + AM + BM = 5 + 5.66 + 5.2 = 15.82 cm.

Answer: The perimeter of the ABM triangle is 15.82 cm.