document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=151452>Question 151452</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1) ABCD is a parallelogram if AB = 2 AD and P is the midpoint of AB.prove that angle CPD = 90. \r<BR>\n" );
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document.write( "2) If the diagonal PR and QS of a parallelogram PQRS are equal, prove that PQRS is a rectangle.\r<BR>\n" );
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document.write( "3) PQRS is a parallelogram. PS is produced to M so that SM = SR and MR is produced to meet PQ produced to N. Prove that QN = QR.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #112187 by </B><A HREF=/tutors/aboutme.mpl?userid=orca><B>orca(409)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=orca><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=112187\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1) ABCD is a parallelogram if AB = 2 AD and P is the midpoint of AB.prove that angle CPD = 90. <BR>\n" );
document.write( "SOLUTION:<BR>\n" );
document.write( "In isosceles triangle APD,<BR>\n" );
document.write( "< APD = < ADP = (180 - < A)/2 = 90 - < A/2<BR>\n" );
document.write( "In isosceles triangle PCB,<BR>\n" );
document.write( "< BPC = < BCP = (180 - < B)/2 = 90 - < B/2<BR>\n" );
document.write( "Next express < DPC in terms of < A and < B.<BR>\n" );
document.write( "< DPC = 180 - < APD - < BPC = 180 - (90 - < A/2) - (90 - < B/2) = < A/2+ < B/2<BR>\n" );
document.write( "= (< A + < B)/2\r<BR>\n" );
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document.write( "Note that in any parallelogram the sum of any two adjacent angles is 180. so:<BR>\n" );
document.write( "< A + < B = 180<BR>\n" );
document.write( "Therefore <BR>\n" );
document.write( "< DPC = (< A + < B)/2 = 180/2 = 90<BR>\n" );
document.write( "ALTERNATIVE SOLUTION:<BR>\n" );
document.write( "First prove that<BR>\n" );
document.write( "PD bisects < D<BR>\n" );
document.write( "PC bisect < C<BR>\n" );
document.write( "(Let Q be the midpoint of CD. As triangle APD and QPD are congruent, so < ADP = < QDP. For the same reason < BCP = < QCP)<BR>\n" );
document.write( "Next<BR>\n" );
document.write( "< PDQ = < D/2<BR>\n" );
document.write( "< PCQ = < C/2<BR>\n" );
document.write( "Thus < PDQ + < PCQ = < D/2 + < C/2 = (< D + < C)/2 = 180/2 = 90.<BR>\n" );
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document.write( "2) If the diagonal PR and QS of a parallelogram PQRS are equal, prove that PQRS is a rectangle.<BR>\n" );
document.write( "SOLUTION:<BR>\n" );
document.write( "We need to prove its angles are 90 degree.<BR>\n" );
document.write( "First prove that triangle PQR and triangle QRS are congruent.(Reason: PQ = SR, QR = QR and PR = QS) <BR>\n" );
document.write( "So < Q = < R<BR>\n" );
document.write( "Next note that < Q + < R = 180.<BR>\n" );
document.write( "As < Q = < R, < Q = < R = 90.<BR>\n" );
document.write( "So PQRS is a rectangle.<BR>\n" );
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document.write( "PQRS is a parallelogram. PS is produced to M so that SM = SR and MR is produced to meet PQ produced to N. Prove that QN = QR. <BR>\n" );
document.write( "SOLUTION:<BR>\n" );
document.write( "We need to prove triangle QNR is an isosceles triangle.<BR>\n" );
document.write( "First show that < N = < SRM and < QRN = < M ( ? )<BR>\n" );
document.write( "Next show that < SRM = < M ( ? )<BR>\n" );
document.write( "Conclusion: < N = < QRN ( ? ), So triangle QNR is isosceles.\r<BR>\n" );
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