2018-II-39

Ans: B
Join $AC$ and $AD$.

Consider $\mathrm{\Delta }ABC$.

$\begin{array}{rcll}\mathrm{\angle }ACB& =& \mathrm{\angle }BAT& \text{(}\mathrm{\angle }\text{s in alt. segment)}\\ \mathrm{\angle }ACB& =& {24}^{\circ }\end{array}$

Also,

$\begin{array}{rcll}\mathrm{\angle }CAE& =& \mathrm{\angle }ABC& \text{(}\mathrm{\angle }\text{s in alt. segment)}\end{array}$

Since $AB=CD\text{(given)}$,

$\begin{array}{rcll}\therefore \stackrel{⏜}{AB}& =& \stackrel{⏜}{CD}& \text{(eq. chords, eq. arcs)}\\ \therefore \mathrm{\angle }CAD& =& \mathrm{\angle }ACB& \text{(arcs and }\mathrm{\angle }\text{s at }{\text{⊙}}^{ce}\text{ in proportion)}\\ \mathrm{\angle }CAD& =& {24}^{\circ }\end{array}$

Consider $\mathrm{\Delta }ADE$.

$\begin{array}{rcll}\mathrm{\angle }ADE& =& \mathrm{\angle }ABC& \text{(ext. }\mathrm{\angle }\text{, cyclic quad.)}\\ \mathrm{\angle }DAE& =& \mathrm{\angle }CAE–\mathrm{\angle }CAD\\ \mathrm{\angle }DAE& =& \mathrm{\angle }ABC–{24}^{\circ }\end{array}$

Hence, we have

$\begin{array}{rcll}\mathrm{\angle }AED+\mathrm{\angle }ADE+\mathrm{\angle }DAE& =& {180}^{\circ }& \text{(}\mathrm{\angle }\text{ sum of }\mathrm{\Delta }\text{)}\\ {72}^{\circ }+\mathrm{\angle }ABC+\mathrm{\angle }ABC–{24}^{\circ }& =& {180}^{\circ }\\ \mathrm{\angle }ABC& =& {66}^{\circ }\end{array}$