Question 1104836: Given : Parallelogram PQRS, line segment PE is perpendicular to line segment SQ, line segment RF is perpendicular to line segment SQ. Prove : line segment SQ is congruent to line segment QF
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
In fact that can't be proved, because SQ is NOT congruent to QF. In fact, clearly QF is part of QS.
However, if you had stated the problem correctly, we can prove that SE is congruent to QF.
If you want to get an answer to your problem, then it is worth the effort on your part to make sure you ask the question correctly. There are a number of "tutors" who answer questions on this web site who would simply say "they are not congruent" and leave their response at that -- making you re-submit the question.
There are undoubtedly many different ways to do this proof. The thing that occurred to me first was to use the areas of triangles PQS and RQS.
We know those triangles are congruent by SSS, so their areas are equal.
Since SQ can be used as the base of both triangles, the altitudes are PE and RF.
Since the areas of the two triangles are the same and the lengths of the bases are the same, the altitudes are the same length.
Then triangles RFQ and PES are congruent by hypotenuse-leg; and therefore QF is congruent to SE.
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