Question

In the given figure:
$AC=BD;\angle BAD={110}^{\circ }$ and $\angle ACB={74}^{\circ }$.
then : $BC>CD$

• A. True
• B. False

Answer 1

The correct option is A True

Given$:\angle ACB={74}^{\circ }$ .....$\left(1\right)$

$\angle ACB+\angle ACD={180}^{\circ }$ since $BCD$ is a straight line.

$\Rightarrow {74}^{\circ }+\angle ACD={180}^{\circ }$

$\Rightarrow \angle ACD={180}^{\circ }-{74}^{\circ }={106}^{\circ }$ .....$\left(2\right)$

In $△ACD,$

$\angle ACD+\angle ADC+\angle CAD={180}^{\circ }$

Given that $AC=CD$

$\angle ADC=\angle CAD$

$\Rightarrow {106}^{\circ }+\angle CAD+\angle CAD={180}^{\circ }$ from $\left(2\right)$

$\Rightarrow 2\angle CAD={180}^{\circ }-{106}^{\circ }={74}^{\circ }$

$\Rightarrow \angle CAD={37}^{\circ }=\angle ADC$ ......$\left(3\right)$

Now,Given$:\angle BAD={110}^{\circ }$

$\angle BAC+\angle CAD={110}^{\circ }$

$\angle BAC+{37}^{\circ }={110}^{\circ }$

$\angle BAC={110}^{\circ }-{37}^{\circ }={73}^{\circ }$ ......$\left(4\right)$

In $△ABC,$

$\angle B+\angle BAC+\angle ACB={180}^{\circ }$ from $\left(1\right)$ and $\left(4\right)$

$\angle B+{73}^{\circ }+{74}^{\circ }={180}^{\circ }$

$\angle B+{147}^{\circ }={180}^{\circ }$

$\angle B={180}^{\circ }-{147}^{\circ }={33}^{\circ }$ ........$\left(5\right)$

$\therefore \angle BAC>\angle B$ from $\left(4\right)$ and $\left(5\right)$

$\Rightarrow BC>AC$

But $AC=CD$(given)

$\Rightarrow BC>CD$

Hence the given statement is true.