In an isosceles triangle ABC (ab = bc), the angle at the vertex b is 120 °, Cm is a bisector, am = 14cm.

| September 6, 2021 | education

In an isosceles triangle ABC (ab = bc), the angle at the vertex b is 120 °, Cm is a bisector, am = 14cm. then the distance from point m to line bc is?

Let the length of the segment BM = X cm, then the length AB = (14 + X) cm.

Since the triangle ABC is isosceles, then BC = AB = (14 + X) cm.

The angles of the ABC triangle at the base of the AC are equal: (180 – 120) / 2 = 30.

Let’s build the height BK, which is also the bisector and the median, then the triangle BCК is rectangular, in which CК = BC * Cos30 = (14 + X) * √3 / 2.

Then AC = 2 * CK = (14 + X) * √3.

By the property of the CM bisector:

AC / AM = BC / BM.

((14 + X) * √3) / 14 = (14 + X) / X.

√3 / 14 = 1 / X.

X = BM = 14 / √3 cm.

Triangle BHM is rectangular in which the angle HBK = (180 – 120) = 60.

Then МН = ВМ * Sin60 = (14 / √3) * (√3 / 2) = 7 cm.

Answer: From point M to BC 7 cm.



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