Through the vertex A of the parallelogram ABCD, a straight line is drawn intersecting the side BC at point E

| April 9, 2021 | education

Through the vertex A of the parallelogram ABCD, a straight line is drawn intersecting the side BC at point E and the continuation of the side DS at point M. Prove that triangle ABE is similar to triangle EMC. Find BE if BC = 10cm, AB = 8cm, CM = 12cm.

In triangles ABE and EMC, the angles CEM and AED are equal as the vertical angles formed by the intersection of lines BC and AM.

The angle ECM and ABE is equal to both the crossed angles formed by the intersection of the secant DC of the parallel lines DM and AB.

Then the triangles ABE and EMC are similar in two angles, as required.

Let the length of the segment BE = X cm, then CE = (10 – X) cm.

From the similarity of triangles: AB / BE = CM / CE.

8 / X = 12 / (10 – X).

12 * X = 80 – 8 * X.

20 * X = 80.

X = BE = 80/20 = 4 cm.

Answer: The length of the segment BE is 4 cm.



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