Question

Question 135
In a rectangle ABCD, AB = 25 cm and BC = 15 cm. In what ratio, does the bisector of $\angle C$ divide AB?

Answer 1

Given, AB = 25 cm and BC = 15 cm
Now, in rectangle ABCD,

CO is the bisector of $\angle C$ and it divides AB.
$\therefore \angle OCB=\angle OCD={45}^{\circ }$

In $\Delta OCB$,
$\angle CBO+\angle OCB+\angle COB={180}^{\circ }$ [angle sum property of triangle]
${90}^{\circ }+{45}^{\circ }+\angle COB={180}^{\circ }\angle COB={180}^{\circ }-({90}^{\circ }+{45}^{\circ })\angle COB={180}^{\circ }-{135}^{\circ }={45}^{\circ }$

Now, in $\Delta OCB$,
$\angle OCB=\angle COB$
Then, OB = BC
$\Rightarrow OB=15cm$
$\Rightarrow AO=25-OB=25-15=10cm$
CO divides AB in the ratio AO : OB
$\therefore$ CO divides AB in the ratio: $10:15=2:3$