document.write( "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 \n" ); document.write( "

Algebra.Com's Answer #719563 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "In fact that can't be proved, because SQ is NOT congruent to QF. In fact, clearly QF is part of QS.

\n" ); document.write( "However, if you had stated the problem correctly, we can prove that SE is congruent to QF.

\n" ); document.write( "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.

\n" ); document.write( "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.

\n" ); document.write( "We know those triangles are congruent by SSS, so their areas are equal.
\n" ); document.write( "Since SQ can be used as the base of both triangles, the altitudes are PE and RF.
\n" ); document.write( "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.
\n" ); document.write( "Then triangles RFQ and PES are congruent by hypotenuse-leg; and therefore QF is congruent to SE. \n" ); document.write( "

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