Question

$In\Delta ABC,\text{side AB is produced to D such that BD = BC.}If\angle A={70}^{\circ }and\angle B={60}^{\circ },\text{prove that (i) AD > CD (ii) AD > AC.}$

Answer 1

Solution

n $△CBA$ CBD is an exterior angle.
i.e., $\angle CBA+\angle CBD={180}^0$

⇒${60}^0+\angle CBD={180}^0\Rightarrow \angle CBD={120}^0$
$△BCDisisoscelesandBC=BD$
$Let\angle BCD=\angle BDC=x^0$
In \(\triangle CBD, we have:⇒\angle BCD+\angle CBD+\angle CDB=180°⇒x+120°+x=180⇒2x=60°⇒x=30°∴\angle BCD=\angle BDC=30°
In \(\triangle ADC, \angle C=\angle ACB + \angle BCD = 50°+30°=80°
\angle A=70°and \angle D=30°∴\angle C>\angle A⇒AD>CD
...(1)
Also, \angle C>\angle D⇒AD>AC ...(2)