Through an arbitrary point of the base of an isosceles triangle, straight lines are drawn parallel to the lateral

| January 7, 2021 | education

Through an arbitrary point of the base of an isosceles triangle, straight lines are drawn parallel to the lateral sides of the triangle. Prove that the perimeter of the resulting quadrilateral is equal to the sum of the sides of this triangle.

Since KD is parallel to the BC, and the DM is parallel to AB, then the quadrangle of the DKВM is a parallelogram, and therefore KD = ВM, KВ = DM.
The AKD triangle is similar to the ABC triangle in two angles, and since the ABC triangle, by condition, is equilateral, the AKD triangle is equilateral, AK = KD.
The СDM triangle is similar to the ABC triangle in two angles, and since the ABC triangle, by condition, is equilateral, the СMD triangle is equilateral, CM = DM.
Then the perimeter of the DKВM will be equal to: R = (KD + ВK) + (ВM + DM) = (AK + ВK) + (ВM + CM) = AB + СВ.
Which is what was required to be proved.



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