The bisectors of angles A and D of parallelogram ABCD meet at point M lying on side BC. Beam DM intersects line AB

| April 12, 2021 | education

The bisectors of angles A and D of parallelogram ABCD meet at point M lying on side BC. Beam DM intersects line AB at point N. Find the perimeter of parallelogram ABCD if AN = 10 cm.

By the property of the bisectors of a parallelogram, they cut off isosceles triangles from the lateral sides, therefore, triangles ABM and DCM are isosceles. Then AB = BM, CD = CM, and since the sides of the parallelogram are equal, AB = CD = BM = CM, and BC = BM + CM = 2 * AB.

Let side AB = X cm, then BM = X cm, BC = AD = 2 * X cm, and BN = (10 – X) cm.

Let us prove that triangles АNД and BNM are similar.

The angle N in triangles is common, the angle NAD is equal to the angle NBM as the corresponding angles at the intersection of parallel lines AD and BC of the secant AN. Then triangles АND and BNM are similar in the first sign of similarity – two angles.

Then:

AD / BM = AN / BN.

2 * X / X = 10 / (10 – X).

10 * X = 20 * X – 2 * X ^ 2.

10 = 20 – 2 * H.

2 * X = 10.

X = 10/2 = 5 cm.

AB = CD = 5 cm.

BC = AD = 2 * 5 = 10 cm.

Determine the perimeter of the parallelogram.

P = 2 * (AB + BC) = 2 * (5 + 10) = 30 cm.

Answer: the perimeter is 30 cm.



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