Question

In $\Delta ABC,\text{AB is produced to D so that}BD=BC.if\angle B={60}^{\circ }and\angle A={70}^{\circ },provethat:\left(i\right)AD>CD\left(ii\right)AD>AC$

Answer 1

Given : In $\Delta ABC,\text{side BC is produced to D such that BD=BC}$

$\angle A={70}^{\circ }and\angle B={60}^{\circ }$

To prove :

(i) $AD>CD\left(ii\right)AD>AC$

Proof : In $\Delta ABC,$

$\angle A={70}^{\circ },\angle B={60}^{\circ }$

But Ext, $\angle CBD+\angle CBA={180}^{\circ }$

(Linear pair)

$\angle CBD+{60}^{\circ }={180}^{\circ }$

$\Rightarrow \angle CBD={180}^{\circ }-{60}^{\circ }={120}^{\circ }$

But in $\Delta BCD,$

BD = BC

$\therefore \angle D=\angle BCD$

But $\angle D+\angle BCD={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

$\therefore \angle D=\angle BCD=\frac{{60}^{\circ }}{2}={30}^{\circ }$

and in $\Delta ABC,$

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

$\Rightarrow {70}^{\circ }+{60}^{\circ }+\angle ={180}^{\circ }$

$\Rightarrow {130}^{\circ }+\angle C={180}^{\circ }$

$\therefore \angle C={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

Now $\angle ACD=\angle ACB+\angle BCD$

=${50}^{\circ }+{30}^{\circ }={80}^{\circ }$

(i) Now in $\Delta ACB,$

$\angle ACD={80}^{\circ }and\angle A={70}^{\circ }$

$\therefore SideAD>CD$

(Greater angle has greatest side opposite to it)

(ii) $\because \angle ACD={80}^{\circ }and\angle D={30}^{\circ }$

$\therefore AD>AC$