document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=31174>Question 31174</A>: <I><SPAN STYLE=\"word-break: break-all;\"> This problem just confuses me to no end. Please help.<BR>\n" );
document.write( "M is the midpoint of AB and P is any point in AM. Line MD meeting BC at D is parallel to PC. Prove that triangle BPD has half the area of triangle ABC.\r<BR>\n" );
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document.write( "This que<BR>\n" );
document.write( "In triangle ABC, stion is so hard. I really hope you can find a solution! Thank you!!! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #18011 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=18011\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This problem just confuses me to no end. Please help.<BR>\n" );
document.write( "In triangle ABC, M is the midpoint of AB and P is any point in AM. Line MD meeting BC at D is parallel to PC. Prove that triangle BPD has half the area of triangle ABC. <BR>\n" );
document.write( "OK ...P IS ANY POINT ON AM.LET <BR>\n" );
document.write( "MP=X*MA=X*AB/2..SINCE M IS MIDPOINT OF AB.<BR>\n" );
document.write( "IN TRIANGLE BPC,WE HAVE MD || PC.HENCE<BR>\n" );
document.write( "BD/DC=BM/MP=(AB/2)/(X*AB/2)=1/X<BR>\n" );
document.write( "BD/DC=1/X.....OR....BD/(BD+DC)=1/(1+X)<BR>\n" );
document.write( "BD/BC=1/(1+X)...OR.....BD=BC/(1+X)................................I<BR>\n" );
document.write( "SIMILARLY FROM BM/MP = 1/X.....OR.......BM/(BM+MP)=1(1+X)<BR>\n" );
document.write( "BM/BP=1/(1+X)<BR>\n" );
document.write( "(AB/2)/BP=1/(1+X)......OR.....BP=AB*(1+X)/2.....................II<BR>\n" );
document.write( "NOW AREA OF TRIANGLE = (1/2)*BASE*ALTITUDE<BR>\n" );
document.write( "AREA OF TRIANGLE ABC = A(ABC)SAY =(1/2)BC*AB*SIN(B)..................III<BR>\n" );
document.write( "SINCE ALTITUDE =AB*SIN(B).......<BR>\n" );
document.write( "SIMILARLY.....<BR>\n" );
document.write( "AREA OF TRIANGLE BPD=A(BPD) SAY =(1/2)BD*BP*SIN(B)........................IV<BR>\n" );
document.write( "DIVIDING EQN.III/EQN.IV.....WE GET.....<BR>\n" );
document.write( "A(ABC)/A(BPD)=BC*AB/BD*BP..............................................V<BR>\n" );
document.write( "BUT BD=BC/(1+X) AND BP =AB*(1+X)/2...SO<BR>\n" );
document.write( "BD*BP=BC*AB*(1+X)/2(1+X)=BC*AB/2...SUBSTITUTING IN EQN.V WE GET <BR>\n" );
document.write( "A(ABC)/A(BPD)=BC*AB/(BC*AB/2)=2.....PROVED </SPAN>\n" );
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