In fig. 5.11. Let points A and B be on one side of a line l. Draw Perpendicular AD and perpendicular BE to line L, C be the mid point of AB. Prove that seg CD $≅$ seg CE.

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Solution

Disclaimer : This question seems to be incomplete ,
It has to be given that M is mid point of seg DE

We know that a segment joining the midpoint of non-parallel sides of a trapezium is parallel to its parallel sides.
Therefore, AD $∥$ CM $∥$ BE

We know that if we draw a line parallel to the parallel sides of a trapezium, then it divides the non-parallel sides in the same ratio.
Here,
$\frac{DM}{ME} = \frac{AC}{CB} Or, DM = ME (∵ AC = CB)$
or, seg DM $≅$ seg ME

Now, consider $△$ CMD and $△$ CME
seg MD $≅$ seg ME (From above)
$∠$ CMD = $∠$ CME (90 $°$ each )
seg CM $≅$ seg CM (Common arm)

By SAS congruency, $△$ CMD $≅$ $△$ CME
seg CD $≅$ seg CE (by c.s.c.t.)

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In fig. 5.11. Let points A and B be on one side of a line l. Draw Perpendicular AD and perpendicular BE to line L, C be the mid point of AB. Prove that seg CD ≅ seg CE.

In fig. 5.11. Let points A and B be on one side of a line l. Draw Perpendicular AD and perpendicular BE to line L, C be the mid point of AB. Prove that seg CD $≅$ seg CE.