2017-II-18

Ans: A

Consider $\mathrm{\Delta }ACD$,

$\begin{array}{rcl}\mathrm{\angle }ADC& =& {180}^{\circ }–{124}^{\circ }\text{(adj. }\mathrm{\angle }\text{s on a st. line)}\\ \mathrm{\angle }ADC& =& {56}^{\circ }\end{array}$

Since $AC=AD$, then we have

$\begin{array}{rcl}\mathrm{\angle }ACD& =& \mathrm{\angle }ADC\text{(base }\mathrm{\angle }\text{, isos. }\mathrm{\Delta }\text{)}\\ \mathrm{\angle }ACD& =& {56}^{\circ }\end{array}$

Therefore, we have

$\begin{array}{rcl}\mathrm{\angle }CAD& =& {180}^{\circ }–\mathrm{\angle }ADC–\mathrm{\angle }ACD\text{(}\mathrm{\angle }\text{ sum of }\mathrm{\Delta }\text{)}\\ \mathrm{\angle }CAD& =& {180}^{\circ }–{56}^{\circ }–{56}^{\circ }\\ \mathrm{\angle }CAD& =& {68}^{\circ }\end{array}$

Since $AE\text{//}BC$, then we have

$\begin{array}{rcl}\mathrm{\angle }BCA& =& \mathrm{\angle }CAD\text{(alt. }\mathrm{\angle }\text{s, }AE\text{//}BC\text{)}\\ \mathrm{\angle }BCA& =& {68}^{\circ }\end{array}$

Consider $\mathrm{\Delta }ABC$. Since $AB=BC$, we have

$\begin{array}{rcl}\mathrm{\angle }BAC& =& \mathrm{\angle }BCA\text{(base }\mathrm{\angle }\text{, isos. }\mathrm{\Delta }\text{)}\\ \mathrm{\angle }BAC& =& {68}^{\circ }\end{array}$

Therefore, we have

$\begin{array}{rcl}\mathrm{\angle }ABC& =& {180}^{\circ }–\mathrm{\angle }BAC–\mathrm{\angle }BCA\text{(}\mathrm{\angle }\text{ sum of }\mathrm{\Delta }\text{)}\\ \mathrm{\angle }ABC& =& {180}^{\circ }–{68}^{\circ }–{68}^{\circ }\\ \mathrm{\angle }ABC& =& {44}^{\circ }\end{array}$