SOLUTION: 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.

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.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
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.
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
THEN BD/CD IS NEGATIVE ETC
LET ABC BE THE TRIANGLE.BP AND CP ARE EXTERNAL BISECTORS MEETING AT P.JOIN AP
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.
HENCE USING CEVAS THEOREM (BD/DC)(CE/EA)(AF/FB)=1.........1
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
CE/EA=BC/BA..........BE BEING BISECTOR OF ANGLE B.
AF/FB=CA/CB..........CF BEING BISECTOR OF ANGLE C.
HENCE FROM EQN.1
(BD/DC)(BC/BA)(CA/CB)=1
BD/DC=BA/AC..........
HENCE BD IS THE BISECTOR OF ANGLE B BY CONNVERSE OF THE THEOREM STATED ABOVE.
HENCE 2 EXTERNAL BISECTORS OF 2 ANGLES IN A TRIANGLE AND INTERNAL BISECTOR OF THIRD ANGLE ARE CONCURRENT
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