document.write( "Question 1064595: A circle is drawn that intersects all three sides of triangle PQR as shown below. Prove that if AB = CD = EF, then the center of the circle is the incenter of triangle PQR. Sorry for posting a link without any picture :).\r
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Algebra.Com's Answer #679668 by ikleyn(52908)\"\" \"About 
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\n" ); document.write( " A circle is drawn that intersects all three sides of triangle PQR as shown below. Prove that if AB = CD = EF,
\n" ); document.write( "then the center of the circle is the incenter of triangle PQR.
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document.write( "Let d be the common length |AB| = |CD| = |EF| = d.\r\n" );
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document.write( "Let \"O\" be the center of the given circle.\r\n" );
document.write( "Connect this point O with the intersection points C and D by the radii OC and OD.\r\n" );
document.write( "The triangle COD is isosceles (its sides OC and OD are the radii !). \r\n" );
document.write( "Draw the perpendicular OM from O to CD. Then OM is the altitude in the isosceles triangle COD; hence, OM is the median, too.\r\n" );
document.write( "It means that CM = DM = \"d%2F2\".\r\n" );
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document.write( "Calculate the length of OM from the right-angled triangle MOD:\r\n" );
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document.write( "|OM| = \"sqrt%28R%5E2+-+%28d%2F2%29%5E2%29\".\r\n" );
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document.write( "If you make similar construction/drawing and calculations for two other triangles OAB and OEF, you will see that\r\n" );
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document.write( "their altitudes are expressed by the same formula, and, hence, have THE SAME length.\r\n" );
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document.write( "Thus the point O is equidistant from the three sides of the given triangle.\r\n" );
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document.write( "Hence, the point O is the center of the inscribed circle to the given triangle PQR.\r\n" );
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