Question
In $\triangle\text{ABC,}$ side AB is produced to D such that BD = BC. If $\angle\text{B}=60^{\circ},$ and $\angle\text{B}=60^{\circ},$ prove that:
- AD > CD and
- AD > AC.
✓
Answer
In triangle CBA, CBD is an exterior angle.
i.e., $\angle\text{CBA}+\angle\text{CBD}=180^{\circ}$
$\Rightarrow60^{\circ}+\angle\text{CBD}=180^{\circ}$
$\Rightarrow\angle\text{CBD}=120^{\circ}$
Triangle BCD is isosceles and BC = BD.
Let $\angle\text{BCD}=\angle\text{BCD}=\text{x}^{\circ}$
In $\triangle\text{CBD},$ we have:
$\Rightarrow\angle\text{BCD}+\angle\text{CBD}+\angle\text{CDB}=180^{\circ}$
$\Rightarrow\text{x}+120^{\circ}+\text{x}=180$
$\Rightarrow2\text{x}=60^{\circ}$
$\Rightarrow\text{x}=30^{\circ}$
$\therefore\angle\text{BCD}=\angle\text{BDC}=30^{\circ}$
In triangle ABC, $\angle\text{C}=\angle\text{ACB}+\angle\text{BCD}=50^{\circ}+30^{\circ}=80^{\circ}$
$\angle\text{A}=70^{\circ}$
and $\angle\text{D}=30^{\circ}$
$\therefore\angle\text{C}>\angle\text{A}$
$\Rightarrow\text{AD}>\text{CD}...(1)$
Also, $\angle\text{C}>\angle\text{D}$
$\Rightarrow\text{AD}>\text{AC}...(2)$
i.e., $\angle\text{CBA}+\angle\text{CBD}=180^{\circ}$
$\Rightarrow60^{\circ}+\angle\text{CBD}=180^{\circ}$
$\Rightarrow\angle\text{CBD}=120^{\circ}$
Triangle BCD is isosceles and BC = BD.
Let $\angle\text{BCD}=\angle\text{BCD}=\text{x}^{\circ}$
In $\triangle\text{CBD},$ we have:
$\Rightarrow\angle\text{BCD}+\angle\text{CBD}+\angle\text{CDB}=180^{\circ}$
$\Rightarrow\text{x}+120^{\circ}+\text{x}=180$
$\Rightarrow2\text{x}=60^{\circ}$
$\Rightarrow\text{x}=30^{\circ}$
$\therefore\angle\text{BCD}=\angle\text{BDC}=30^{\circ}$
In triangle ABC, $\angle\text{C}=\angle\text{ACB}+\angle\text{BCD}=50^{\circ}+30^{\circ}=80^{\circ}$
$\angle\text{A}=70^{\circ}$
and $\angle\text{D}=30^{\circ}$
$\therefore\angle\text{C}>\angle\text{A}$
$\Rightarrow\text{AD}>\text{CD}...(1)$
Also, $\angle\text{C}>\angle\text{D}$
$\Rightarrow\text{AD}>\text{AC}...(2)$
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