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The bisector of interior `angleA of triangleABC` meets BC in D, and the bisector of exterior angle `angleA `meets BC produced in E. prove that `(BD)/(BE)= (CD)/(CE)`

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Knowledge Check

  • The bisector of angleA of DeltaABC cuts BC at D and the circumcircle of the triangle at E. then

    A
    AB:AC=BD:DC
    B
    AD:AC=AE:AB
    C
    AB:AD=AC:AE
    D
    AB:AB=AE:AC

  • The bisector of the exterior angleA of DeltaABC intersects the side BC produced to D. Here CF is parallel to AD.

    A
    `(AB)/(AC)=(BD)/(CD)`
    B
    `(AB)/(AC)=(CD)/(BD)`
    C
    `(AB)/(AC)=(BC)/(CD)`
    D
    none of these

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