Angle Bisector Theorem Proof

Exterior angle bisector theorem proof :

Statement :   THE EXTERNAL  BISECTOR OF AN ANGLE OF A TRIANGLE  DIVIDES THE OPPOSITE SIDE  EXTERNALLY  IN THE RATIO OF SIDES CONTAINING THE ANGLE.

GIVEN :  a triangle ABC   in which AD  is the bi sector of  the exterior angle A  and intersects  BC produced in D

REQUIRED TO PROVE : BD /CD = AB /AC

CONSTRUCTION ; draw   CE parallel to  DA  meeting AB  in   E

PROOF ;  since  CE parallel to DA  and  AC   intersects  them

angle  CAD =  angle ACE   (alternate angles)

also CE is parallel to  DA  and    BK  intersects them

so   angle KAD = angle AEC   (corresponding angles)

therfore    angle  ACE  = angle AEC

thus in  triangle ACE     angle  ACE = angle AEC

SO   AE = AC (  sides opposite to equal angles are equal )

in triangle BAD   EC is parallel to AD

THERE FORE        IF A LINE DRAWN PARALLEL TO ONE SIDE OF A TRIANGLE IT DIVIDES THE OTHER TWO SIDES IN THE SAME RATIO.                     BD/CD = BA/EA

BD/CD=AB/AE        reason  BA=AB  AND     EA= AE

BD/CD = AB/AC      reason  AE= AC

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Exterior angle bisector theorem proof : Example

in triangle ABC  AE is  the bisector of exterior angleCAD   meeting  BC  produced  in E  if  AB = 10 CM  AC =6 CE BC = 12 CE  find CE ?

solution;     according to exterior angle bisector theorem   BE / CE= AB/AC

let  CE = X

(  12+X ) / X =  10 / 6  CROSS MULTIPLECATION

5X = 36 + 3X

5X - 3X = 36

2X = 36

X=  36/2

X = 18

CE =  18 cm

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ASSIGNMENT

AE is the bisector of the exterior angle CAD  meeting BC  produced  in E  if AB = 12 CM AC = 6 CM  AND BC = 14 cm  find CE ?

LET   CE =  X

BE / CE  =AB / AC  (according to external angle bisector theorem )

14 + x /  x  = 12 / 6

14 + x  / x =    2

by  cross multiplection

2x - x =  14

x =   14

CE = 14