Question

In $△DEF$ shown below, points $A,B,C$ are taken on $DE,DF$ and $EF$ respectively such that $EC=AC$ and $CF=BC$, $\angle D={40}^o$. What is $\angle ACB$ in degrees?

• A. ${140}^o$
• B. ${70}^o$
• C. ${100}^o$
• D. ${80}^o$

Answer 1

Answer: (C)

${100}^o$

In $△DEF$
$\angle EDF+\angle DEF+\angle DFE={180}^o$
$\Rightarrow$ $\angle DEF+\angle DFE={180}^o-\angle EDF$
$\Rightarrow$ $\angle DEF+\angle DFE={180}^o-{40}^o={140}^o......\left(i\right)$
In $△BCF$, $CB=CF$
$\Rightarrow$ $\angle CFB=\angle CBF$ (Isos.$△CBF$)
or $\angle EFD=\angle CBF.......\left(ii\right)$
In $△AEC,CA=CE\Rightarrow \angle CEA=\angle CAE$ (Isos. $△$ prop.)$......\left(iii\right)$
or $\angle DEF=\angle CAE$
$\therefore$ from $\left(i\right),\left(ii\right),\left(iii\right)$
$\angle CAE+\angle CBF={140}^o$
${180}^o-\angle CAD+{180}^o-\angle CBD={140}^o$
$\Rightarrow$ $\angle CAD+\angle CBD={360}^o-{140}^o={220}^o$
In quad, $DACB$
$\angle ADB+\angle CAD+\angle CBD+\angle ACB={360}^o$
$\Rightarrow$ ${40}^o+{220}^o+\angle ACB={360}^o$
$\Rightarrow$ $\angle ACB={360}^o-{260}^o={100}^o$