Through an arbitrary point D, the base AC of an isosceles triangle ABC, straight lines are drawn that are parallel

| May 23, 2021 | education

Through an arbitrary point D, the base AC of an isosceles triangle ABC, straight lines are drawn that are parallel to the lateral sides of the triangle and intersect them at points M and M. Find the perimeter BMDN if AB = 10 cm.

By condition, triangle ABC is isosceles, AB = BC. Then the angle BAC = BCA.

Consider triangles АМD and ABC, in which the angle A is common, the angle АDM is equal to the angle BCA, as the corresponding angles at the intersection of parallel straight lines BC and MD, and, accordingly, is equal to the angle MAD. The AMD triangle is isosceles at the same angles at the base of AD.

So AM = MD.

Consider triangles CND and CBA, for which the angle C is common, the angle NDC is equal to the angle BAC, as the corresponding angles at the intersection of parallel lines AB and ND, and, accordingly, is equal to the angle DCF. The CND triangle is isosceles at the same angles at the base of the CD.

Hence NC = ND.

Perimeter MBND = (MB + MD) + (BN + ND) = (MB + AM) + (BN + NC) = AB + BC = 2 * AB = 2 * 10 = 20 cm.

Answer: Рmbnd = 20 cm.



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