Question

State true or false:

In $\Delta ABC,$ $\angle ABC$ is equal to twice $\angle ACB$, and the bisector of $\angle ABC$ meets the opposite side at a point P. Then, $CB:BA$ $=$ $CP:PA$.

• A. True
• B. False

Answer 1

The correct option is A True

Given, in $△ABC,\angle ABC=2\angle ACB$

Let $\angle ACB=\theta$, then $\angle ABC=2\theta$

Now, $\angle A+\angle B+\angle C=180$

$\therefore \angle BAC=180-3\theta$

BP is angular bisector of $\angle ABC$.

Let us split them into different triangles.

$\angle APB+\angle ABP+\angle BAP=180°$ ...(triangle law )

$\theta +180-3\theta +\angle BPA=180°$

$\angle BPA=2\theta$

As in $△BPC$, two angles are equal.

we know angles opposites to sides are equal and vice versa

$\Rightarrow BP=CP\to \left(i\right)$

Now compare $△ABC$ and $△ABP$

$\angle BAP=\angle BAC$

$\angle ACB=\angle ABP$

$\angle ABC=\angle APB$

By AAA axiom, $ABC\sim △APB$

$\Rightarrow \frac{AB}{AP}=\frac{BC}{PB}=\frac{AC}{AB}\Rightarrow \frac{BC}{AB}=\frac{PB}{AP}\to \left(2\right)$

$\Rightarrow \frac{BC}{AB}=\frac{CB}{AP}$ ...from (i)