# Segments AE and BD intersect at point C, AB = BC, CD = DE. Triangles ABC and CDE are isosceles. DEC angle = 47.

**Segments AE and BD intersect at point C, AB = BC, CD = DE. Triangles ABC and CDE are isosceles. DEC angle = 47. points K and M were marked on the segments EC and ED so that EK = AC EM = BC. Prove that triangle ABC = MEK.**

Consider triangles ABC and MEK.

AC = EK (by condition)

EM = BC (by condition)

the angle DCE is the angle MEK, so the angle MEK = 47. Since the DCE triangle is isosceles by the condition, the angles at the base are equal by the property of the isosceles triangle, so the angle DEC is equal to the angle DCE, that is, the angle DCE is 47. The angle DCE is angle ACB, since they are vertical, that is, angle ACB = 47. We get that the angle ACB is equal to the angle DEC. So we have

AC = EK (by condition)

EM = BC (by condition)

angle ACB = angle DEC = 47

This means that by the first sign, the triangles ABC and MEK are equal.