Question
From the following figure, prove that: $A B>C D$.
✓
Answer
In $\triangle ABC,$
$AB = AC \dots...[$ Given $]$
$\therefore \angle ACB = \angle B\dots ...[$ angles opposite to equal sides are equal $]$
$\angle B = 70^\circ\dots ...[$ Given $]$
$\Rightarrow \angle ACB = 70^\circ\dots ...(i)$
Now$,$
$\angle ACB +\angle ACD = 180^\circ\dots ...[\text{BCD}$ is a straight line$]$
$\Rightarrow 70^\circ + \angle ACD = 180^\circ $
$\Rightarrow \angle ACD = 110^\circ \dots...(ii)$
In $\triangle ACD,$
$\angle CAD + \angle ACD + \angle D = 180^\circ $
$\Rightarrow \angle CAD + 110^\circ + \angle D = 180^\circ\dots ...[$ From $(ii) ]$
$\Rightarrow \angle CAD + \angle D = 70^\circ $
But $\angle D = 40^\circ\dots ...[$ Given $]$
$\Rightarrow \angle CAD + 40^\circ = 70^\circ $
$\Rightarrow \angle CAD = 30^\circ\dots …(iii)$
In $\triangle ACD,$
$\angle ACD = 110^\circ\dots ...[$ From $(ii) ]$
$\angle CAD = 30^\circ\dots ...[$ From $(iii) ]$
$\angle D = 40^\circ \dots...[$ Given $]$
$\therefore D > \angle CAD$
$\Rightarrow AC > CD\dots ....[$Greater angle has greater side opposite to it$]$
Also$,$
$AB = AC\dots ...[$ Given $]$
Therefore$, AB > CD$
$AB = AC \dots...[$ Given $]$
$\therefore \angle ACB = \angle B\dots ...[$ angles opposite to equal sides are equal $]$
$\angle B = 70^\circ\dots ...[$ Given $]$
$\Rightarrow \angle ACB = 70^\circ\dots ...(i)$
Now$,$
$\angle ACB +\angle ACD = 180^\circ\dots ...[\text{BCD}$ is a straight line$]$
$\Rightarrow 70^\circ + \angle ACD = 180^\circ $
$\Rightarrow \angle ACD = 110^\circ \dots...(ii)$
In $\triangle ACD,$
$\angle CAD + \angle ACD + \angle D = 180^\circ $
$\Rightarrow \angle CAD + 110^\circ + \angle D = 180^\circ\dots ...[$ From $(ii) ]$
$\Rightarrow \angle CAD + \angle D = 70^\circ $
But $\angle D = 40^\circ\dots ...[$ Given $]$
$\Rightarrow \angle CAD + 40^\circ = 70^\circ $
$\Rightarrow \angle CAD = 30^\circ\dots …(iii)$
In $\triangle ACD,$
$\angle ACD = 110^\circ\dots ...[$ From $(ii) ]$
$\angle CAD = 30^\circ\dots ...[$ From $(iii) ]$
$\angle D = 40^\circ \dots...[$ Given $]$
$\therefore D > \angle CAD$
$\Rightarrow AC > CD\dots ....[$Greater angle has greater side opposite to it$]$
Also$,$
$AB = AC\dots ...[$ Given $]$
Therefore$, AB > CD$
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