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prove that the line joining the midpoints of the non parallel sides of trapezium is parallel to the sides of a trapezium and is half of the sum of their parallel sides
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Proof
Let ABCD be a trapezoid with the bases AB and DC and the mid-line EF Compare the triangles DFC and FBG.
The segments FC and BF are congruent since the point F is the midpoint of the side BC. The angles DFC and BFG are congruent as the vertical angles. The angles DCF and FBG are congruent as the alternate exterior angles at the parallel lines AB and DC and the transverse BC. Hence, the triangles DFC and FBG are congruent in accordance with the ASA-test of congruency of triangles. It implies that the segments DF and GF are congruent as the corresponding sides of the congruent triangles DFC and FBG.
Thus the mid-line EF of the trapezoid ABCD is the straight line segment connecting the midpoints of the triangle AGD.
It is well known fact that the the straight line segment connecting the midpoints of the triangle AGD is parallel to the triangle base AG and its length is half of the length of the triangle base. In our case, the length of the segment EF is half of the length AG : |EF| = 1/2*|AG| = 1/2*(|AB| + |BG|). Since |BG| = |DC| from the triangles congruency, we have |EF| = 1/2*(|AB| + |DC|), or |EF| = 1/2*(a + d), where a and d are the lengths of the trapezoid bases.
Summary The mid-line of a trapezoid is parallel to its bases. The length of the mid-line of a trapezoid is half of the sum of the lengths of its bases.
Let ABCD be a trapezoid with the bases AB and DC and the mid-line EF Compare the triangles DFC and FBG.
The segments FC and BF are congruent since the point F is the midpoint of the side BC. The angles DFC and BFG are congruent as the vertical angles. The angles DCF and FBG are congruent as the alternate exterior angles at the parallel lines AB and DC and the transverse BC. Hence, the triangles DFC and FBG are congruent in accordance with the ASA-test of congruency of triangles. It implies that the segments DF and GF are congruent as the corresponding sides of the congruent triangles DFC and FBG.
Thus the mid-line EF of the trapezoid ABCD is the straight line segment connecting the midpoints of the triangle AGD.
It is well known fact that the the straight line segment connecting the midpoints of the triangle AGD is parallel to the triangle base AG and its length is half of the length of the triangle base. In our case, the length of the segment EF is half of the length AG : |EF| = 1/2*|AG| = 1/2*(|AB| + |BG|). Since |BG| = |DC| from the triangles congruency, we have |EF| = 1/2*(|AB| + |DC|), or |EF| = 1/2*(a + d), where a and d are the lengths of the trapezoid bases.
Summary The mid-line of a trapezoid is parallel to its bases. The length of the mid-line of a trapezoid is half of the sum of the lengths of its bases.
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