using cevas theorem prove that the external bisectors of two

using ceva\'s theorem!!!!! prove that the external bisectors of two angles of a triangle and the internal bisector of the third angle are concurrent.

Solution

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