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The external angle bisector of an angle ...

The external angle bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.

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Given: `DeltaABC` in which the bisector AD of `angleA` meets BC ( orBC produced) in D.
To prove: `(BD)/(DC)=(AB)/(AC)`
Construction: Draw CE||DA meeting BA ( produced if necessary) in E.
Proof: Since CE||DA
`angle1=angle2" " ("alternate angles") ....(1)`
` and angle3= angle4` ( corresponding angles) ....(2)
` But angle1= angle3` (given)....(3)
` angle2 = angle4` [ from (1) , (2) and (3)]
AE= AC ( sides opposite to equal angles are equal) .... (4)
Now, since CE||DA ( construction)
` (BD)/(DC)=(BA)/(AE) ....(5)`
` Rightarrow (BD)/(DC)= (BA)/(AC) [ From (4) and (5)] `

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