In the given figure, $A E$ is the bisector of the exterior $∠ C A D$ meeting $B C$ produced in $E$ . If $A B = 10 c m$ , $A C = 6 c m$ and $B C = 12 c m$ , find $C E$ .

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Solution

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

By exterior angle bisector theorem

$\frac{B E}{C E} = \frac{A B}{A C}$

$\frac{B C + C E}{C E} = \frac{A B}{A C}$

$\frac{12 + x}{x} = \frac{10}{6}$

$\frac{12 + x}{x} = \frac{5}{3}$

$36 + 3 x = 5 x$

$36 = 2 x$

$x = 18$

Hence $C E = 18 c m$

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Similar questions

$A E$ is the bisector of the exterior $∠ C A D$ meeting $B C$ produced in $E$ . If $A B = 10 c m$ , $A C = 6 c m$ and $B C = 12 c m$ , find $C E$ .

△ A B C, A E

∠ A

E

A B = 10 c m, A C = 6 c m

B C = 12 c m

C E

∠ C A D

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