Question

In figure, $ABC$ is a triangle with $\angle B={35}^{\circ },\angle C={65}^{\circ }$ and bisector of $\angle BAC$ meets $BC$ in $X$. Arrange $AX,BX$ and $CX$ in descending order.

Answer 1

Consider $△ABC$

By sum property of a triangle

$\angle A+\angle B+\angle C={180}^{\circ }$

To find $\angle A$

$\angle A={180}^{\circ }-\angle B-\angle C$

By substituting the values

$\angle A={180}^{\circ }-{35}^{\circ }-{65}^{\circ }$

By subtraction

$\angle A={180}^{\circ }-{100}^{\circ }$

$\angle A={80}^{\circ }$

We know that

$\angle BAX=\frac{1}{2}\angle A$

So we get

$\angle BAX=\frac{1}{2}\left({80}^{\circ }\right)$

By division

$\angle BAX={40}^{\circ }$

Consider $△ABX$

It is given that $\angle B={35}^{\circ }$ and $\angle BAX={40}^{\circ }$

By sum property of a triangle

$\angle BAX+\angle BXA+\angle XBA={180}^{\circ }$

To find $\angle BXA$

$\angle BXA={180}^{\circ }-\angle BAX-\angle XBA$

By substituting values

$\angle BXA={180}^{\circ }-{35}^{\circ }-{40}^{\circ }$

By subtraction

$\angle BXA={180}^{\circ }-{75}^{\circ }$

$\angle BXA={105}^{\circ }$

We know that $\angle B$ is the smallest angle and the side opposite to it i.e. $AX$ is the smallest side.

So we get $AX<BX..\left(1\right)$

Consider $△AXC$

$\angle CAX=\frac{1}{2}\angle A$

So we get

$\angle CAX=\frac{1}{2}\left({80}^{\circ }\right)$

By division

$\angle CAX={40}^{\circ }$

By sum property of a triangle

$\angle AXC+\angle CAX+\angle CXA={180}^{\circ }$

To find $\angle AXC$

$\angle AXC={180}^{\circ }-\angle CAX-\angle CXA$

By substituting values

$\angle AXC={180}^{\circ }-{40}^{\circ }-{65}^{\circ }$

So we get

$\angle AXC={180}^{\circ }-{105}^{\circ }$

By subtraction

$\angle AXC={75}^{\circ }$

So we know that $\angle CAX$ is the smallest angle and the side opposite to it i.e. $CX$ is the smallest side.

We get

$CX<AX\left(2\right)$

By considering equation $\left(1\right)$ and $\left(2\right)$

$BX>AX>CX$

Therefore, $BX>AX>CX$ is the descending order.