document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=33691>Question 33691</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A regular Pentagon ABCDE is incribed in circle O. Chords EC and DB intersect at F (not diameter). Chord DB is Extended to G (outside circle), and Tangent GA is drawn.<BR>\n" );
document.write( "find measure of angle BDE<BR>\n" );
document.write( "find measure of angle BFC<BR>\n" );
document.write( "find measure of angle AGD </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #20107 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=20107\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A regular Pentagon ABCDE is incribed in circle O. Chords EC and DB intersect at F (not diameter). Chord DB is Extended to G (outside circle), and Tangent GA is drawn.<BR>\n" );
document.write( "find measure of angle BDE<BR>\n" );
document.write( "find measure of angle BFC<BR>\n" );
document.write( "find measure of angle AGD <BR>\n" );
document.write( "HOPE YOU HAVE A FIGURE OF THE ABOVE..ASSUMING THAT I AM GIVING THE SOLUTION.<BR>\n" );
document.write( "SUM OF ALL INTERIOR ANGLES IN A  PENTAGON =(2*5-4)*90 =540<BR>\n" );
document.write( "SINCE IT IS REGULAR PENTAGON EACH ANGLE =540/5=108<BR>\n" );
document.write( "AS PER GIVEN DATA,A,B,E,D LIE ON A CIRCLE...SO ABED  IS A CYCLIC QUADRILATERAL..HENCE SUM OF OPPOSITE ANGLES=180=ANGLE EAB+ANGLE BDE<BR>\n" );
document.write( "BUT ANGLE EAB=108 AS PROVED ABOVE <BR>\n" );
document.write( "-----------------------------------------<BR>\n" );
document.write( "HENCE ANGLE BDE=180-108=72<BR>\n" );
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document.write( "SIMILARLY FROM CYCLIC QUADRILATERAL ECBA,WE GET ANGLE ECB=72<BR>\n" );
document.write( "NOW IN A REGULAR PENTAGON EACH SIDE SPANS 1/5 OF PERIMETER OF CIRCUM CIRCLE WITH O AS CENTRE.HENCE EACH SIDE SUBTENDS AN ANGLE OF 360/5=72 AT CENTRE AND HALF OF THAT THAT IS 36 AT THE CIRCUMFERENCE..<BR>\n" );
document.write( "HENCE ANGLE ADB=ANGLE DBC =36<BR>\n" );
document.write( "NOW IN TRIANGLE CFB...WE HAVE...<BR>\n" );
document.write( "ANGLE FCB =ANGLE ECB =72..PROVED ABOVE.<BR>\n" );
document.write( "ANGLE FBC=ANGLE DBC=36......PROVED ABOVE<BR>\n" );
document.write( "--------------------------------------------<BR>\n" );
document.write( "HENCE ANGLE BFC=180-72-36=72<BR>\n" );
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document.write( "ABDE IS A CYCLIC QUADRILATERAL...AB IS A CHORD...AG IS A TANGENT TO CIRCUM CIRCLE..HENCE ANGLE BAG = ANGLE IN THE OPPOSITE SEGMENT =ANGLE ADB=36..PROVED ABOVE<BR>\n" );
document.write( "HENCE ANGLE EAG=ANGLE EAB + ANGLE BAG=108+36=144<BR>\n" );
document.write( "IN QUADRILATERALDEAG...WE HAVE<BR>\n" );
document.write( "ANGLE AGD=360-ANGLE EDB-ANGLE EAG-ANGLE DEA<BR>\n" );
document.write( "=360-72-144-108=36<BR>\n" );
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document.write( "HENCE ANGLE AGD=36<BR>\n" );
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