In a rectangular trapezoid ABCD with bases AD = 12 and BC = 8 and an angle of BAD = 90 degrees

In a rectangular trapezoid ABCD with bases AD = 12 and BC = 8 and an angle of BAD = 90 degrees, the large diagonal BD = 13. Diagonals meet at point M. 1) Prove that triangles BMC and DMA are similar. 2) find the perimeter of the ABM triangle.

one)

Angle АМD is equal to angle BMC as vertical angles. The MBC angle is equal to the MDA angle as criss-crossing angles at the intersection of parallel straight lines ВС and АD of the secant ВD. Then the triangles BMC and DMA are similar in two angles. Q.E.D.

2)

From the right-angled triangle ABD, according to the Pythagorean theorem, we determine the length of the leg AB.

AB ^ 2 = BD ^ 2 – AD ^ 2 = 13 ^ 2 – 12 ^ 2 = 169 – 144 = 25.

AB = 5 cm.

We draw from the top C the height CH, which is equal in length to the side AB, and divides the base AD into two segments, AH = BC = 8 cm, DH = AD – AH = 12 – 8 = 4 cm.

From the triangle ACН, by the Pythagorean theorem, we define the hypotenuse AC.

AC ^ 2 = AH ^ 2 + CH ^ 2 = 8 ^ 2 + 5 ^ 2 = 64 + 25 = 89.

AC = 9.4 cm.

Let the segment BM = X cm, then the segment DM = (13 – X) cm, then from the similarity with a triangle it follows:

AD / BM = DM / BM.

12/8 = (13 – X) / X.

12 * X = 104 – 8 * X.

20 * X = 104.

X = BM = 5.2 cm.

Let AM = Y cm, then CM = (9.4 – Y) cm.

BP / BM = AM / CM.

12/8 = Y / (9.4 – Y).

8 * Y = 112.8 – 12 * Y.

20 * Y = 112.8.

Y = AM = 5.64 cm.

The perimeter of the ABM triangle is: = 5 + 5.2 + 5.64 = 15.84 cm.

Answer: The perimeter of the ABM triangle is 15.84 cm.