Question

In the figure, if AC is bisector of $\angle BAD$ such that AB = 3 cm and AC = 5 cm, then CD =

• A. 2 cm
• B. 3 cm
• C. 4 cm
• D. 5 cm

Answer 1

The correct option is C

4 cm

In the figure, AC is the bisector of $\angle BAD,$

AB = 3 cm, AC = 5 cm

$In\Delta ABCand\Delta ADC,$

$AC=AC\text{(Common)}$

$\angle B=\angle D\left(Each{90}^{\circ }\right)$

$\angle BAC=\angle DAC$

($\because \text{AC is the bisector of}\angle A)$

$\therefore \Delta ABC\cong \Delta ADC\text{(AAS axiom)}$

$\therefore BC=CDandAB=AD(c.p.c.t.)$

$\text{Now in right}\Delta ABC,$

$A{C}^2=A{B}^2+B{C}^2$

$\Rightarrow (5{)}^2=(3{)}^2+B{C}^2$

$\Rightarrow 25=9+B{C}^2\Rightarrow B{C}^2=25-9=16=(4{)}^2$

$\therefore BC=4cm$

$ButCD=BC$

$\therefore CD=4cm$