Question

In the triangle $ABC,\angle ACE={130}^o,$ segment $AD=AD=DC$, then find the measure of an $\angle ABC$.

Answer 1

Given $\angle ACE={130}^o$
$\angle ACE+\angle ACD={180}^o$
$\angle ACD={180}^o-{130}^o={50}^o$
Given , $AD=CD$

Hence, $\angle CAD=\angle ACD=50$ (Isosceles triangle Property)
$\angle ADC+\angle CAD+\angle ACD={180}^o$ (Angle sum property)
$\angle ADC+{50}^o+{50}^o={180}^o$
$\angle ADC={80}^{\circ }$
Now, $\angle ADC+\angle ADB=180$ (Angles on a straight line)
$80+\angle ADB={180}^o$
$\angle ADB={100}^{\circ }$
Given, $AD=BD$, thus, $\angle DBA=\angle BAD$
In $△ABD$,
$\angle ADB+\angle DBA+\angle BAD={180}^o$
$100+x+x={180}^o$
$2x={80}^o$
$\therefore x={40}^o$
Hence, $\angle DBA$ or $\angle ABC={40}^{\circ }$