Question

In the figure given below, prove that BC < AC < CD.

Answer 1

In triangle ABC,
AB = AC
$\Rightarrow \angle ABC=\angle ACB$
(angles opposite to equal sides are equal)
$\Rightarrow \angle ABC=\angle ACB={67}^{\circ }$

$\Rightarrow \angle BAC={180}^{\circ }-\angle ABC\angle ACB$
(Angle sum property)
$\Rightarrow \angle BAC={180}^{\circ }-{67}^{\circ }-{67}^{\circ }$

$\Rightarrow \angle BAC={46}^{\circ }$

Since, $\angle BAC<\angle ABC$, we have

BC < AC ----------- (1)

Now, $\angle ACD={180}^{\circ }-{67}^{\circ }={113}^{\circ }$
(linear pair)

Thus, in triangle ACD

$\Rightarrow \angle CAD={180}^{\circ }-\angle ACD\angle ADC$
(Angle sum property)

$\Rightarrow \angle CAD={180}^{\circ }-{113}^{\circ }-{33}^{\circ }={34}^{\circ }$

Since, $\angle ADC<\angle CAD$, we have

AC < CD ----------- (2)

From (1) and (2), we have

BC < AC < CD.