Given:
Δ ABC and AD bisects ∠A, meeting side BC at D.
\[~AB\text{ }=\text{ }3.5\text{ }cm\], \[AC\text{ }=\text{ }4.2\text{ }cm,\] and \[DC\text{ }=\text{ }2.8\text{ }cm\].
Required to find: BD
AD is the bisector of ∠ A meeting side BC at D in Δ ABC
⇒ $\frac{AB}{AC}=\frac{BD}{DC}$
$\frac{3.5}{4.2}=\frac{BD}{2.8}$
$4.2\times BD=3.5\times 2.8$
BD = 7/3$BD=\frac{7}{3}$
\[\therefore BD\text{ }=\text{ }2.3\text{ }cm\]
Given:
In Δ ABC, AD is the bisector of ∠A meeting side BC at D.
\[AB\text{ }=\text{ }10\text{ }cm,\] \[AC\text{ }=\text{ }14\text{ }cm,\] and \[BC\text{ }=\text{ }6\text{ }cm\]
Required to find: BD and DC.
AD is bisector of ∠A
AB/AC = BD/DC (AD is bisector of ∠ A and side BC)
$\frac{10}{14}=\frac{x}{\left( 6-x \right)}$
\[14x\text{ }=\text{ }60\text{ }\text{ }6x\]
\[20x\text{ }=\text{ }60\]
\[x\text{ }=\frac{60}{20}\]
\[\therefore \] \[BD\text{ }=\text{ }3\text{ }cm\] and \[DC\text{ }=\text{ }\left( 6\text{ }\text{ }3 \right)\text{ }=\text{ }3\text{ }cm\]