document.write( "Question 54034: Use Ceva's Theorem to prove that the external bisectors of two angles of a triangle and the internal bisector of the third angle are concurrent. \n" ); document.write( "

Algebra.Com's Answer #36329 by venugopalramana(3286) $\"\"$ $\"About$

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Use Ceva's Theorem to prove that the external bisectors of two angles of a triangle and the internal bisector of the third angle are concurrent.
\n" ); document.write( "I AM GIVING A BROAD PROOF..BUT FOR RIGOROUS STATEMENTS PLEASE USE SIGN CONVENTIONS PROPERLY...THAT IS IF BD/DC IS TAKEN AS POSITIVE...IF
\n" ); document.write( "THEN BD/CD IS NEGATIVE ETC
\n" ); document.write( "LET ABC BE THE TRIANGLE.BP AND CP ARE EXTERNAL BISECTORS MEETING AT P.JOIN AP
\n" ); document.write( "TST....AP IS INTERNAL BISECTOR OF ANGLE A.LET AP MEET BC AT D.LET BP PRODUCED MEET AC AT E AND CP PRODUCED MEET AC AT F.
\n" ); document.write( "HENCE USING CEVAS THEOREM (BD/DC)(CE/EA)(AF/FB)=1.........1
\n" ); document.write( "SINCE THE BISECTOR OF AN ANGLE IN A TRIANGLE MEETS DIVIDES THE OPPOSITE SIDE IN THE RATIO OF ADJACENT SIDES FORMING THE ANGLE,WE HAVE
\n" ); document.write( "CE/EA=BC/BA..........BE BEING BISECTOR OF ANGLE B.
\n" ); document.write( "AF/FB=CA/CB..........CF BEING BISECTOR OF ANGLE C.
\n" ); document.write( "HENCE FROM EQN.1
\n" ); document.write( "(BD/DC)(BC/BA)(CA/CB)=1
\n" ); document.write( "BD/DC=BA/AC..........
\n" ); document.write( "HENCE BD IS THE BISECTOR OF ANGLE B BY CONNVERSE OF THE THEOREM STATED ABOVE.
\n" ); document.write( "HENCE 2 EXTERNAL BISECTORS OF 2 ANGLES IN A TRIANGLE AND INTERNAL BISECTOR OF THIRD ANGLE ARE CONCURRENT \n" ); document.write( "

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