In a $Δ A B C, A D$ is the bisector of $∠ A,$ meeting side $B C$ at $D$ .
If $B D = 2.5 c m, A B = 5 c m$ . and $A C = 4.2 c m$ ,
Find $D C$

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Solution

According to angle bisector theorem,
$\frac{A B}{B D} = \frac{A C}{D C}$
$D C = \frac{A C \times B D}{A B} = \frac{4.2 \times 2.5}{5} = 2.1 c m$

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In a ΔABC,AD is the bisector of ∠A, meeting side BC at D.If BD=2.5cm,AB=5cm. and AC=4.2cm,Find DC

According to angle bisector theorem,ABBD=ACDCDC=AC×BDAB=4.2×2.55=2.1 cm

In a $Δ A B C, A D$ is the bisector of $∠ A,$ meeting side $B C$ at $D$ .
If $B D = 2.5 c m, A B = 5 c m$ . and $A C = 4.2 c m$ ,
Find $D C$

According to angle bisector theorem,
$\frac{A B}{B D} = \frac{A C}{D C}$
$D C = \frac{A C \times B D}{A B} = \frac{4.2 \times 2.5}{5} = 2.1 c m$