2021-II-21

Ans: D

I is true.

$\begin{array}{rcl}\mathrm{\angle }ABC& =& {\displaystyle \frac{1}{8}}[(8-2)\times {180}^{\circ }]\\ \mathrm{\angle }ABC& =& {135}^{\circ }\\ \mathrm{\angle }ABF& =& {\displaystyle \frac{1}{2}}\times \mathrm{\angle }ABC\\ \mathrm{\angle }ABF& =& {67.5}^{\circ }\end{array}$

In $\mathrm{\Delta }AHG$, since $AH=GH$, $\mathrm{\angle }AGH=\mathrm{\angle }GAH$ (base $\mathrm{\angle }$s, isos. $\mathrm{\Delta }$).

$\begin{array}{rcll}\mathrm{\angle }GAH& =& {\displaystyle \frac{1}{2}}({180}^{\circ }-\mathrm{\angle }AHG)& \text{(}\mathrm{\angle }\text{ sum of }\mathrm{\Delta }\text{)}\\ \mathrm{\angle }GAH& =& {\displaystyle \frac{1}{2}}({180}^{\circ }-{135}^{\circ })\\ \mathrm{\angle }GAH& =& {22.5}^{\circ }\end{array}$

Therefore, we have

$\begin{array}{rcl}\mathrm{\angle }BAG& =& \mathrm{\angle }BAH-\mathrm{\angle }GAH\\ \mathrm{\angle }BAG& =& {135}^{\circ }-{22.5}^{\circ }\\ \mathrm{\angle }BAD& =& {112.5}^{\circ }\end{array}$

Hence, we have

$\begin{array}{cl}& \mathrm{\angle }ABF+\mathrm{\angle }BAG\\ =& {67.5}^{\circ }+{112.5}^{\circ }\\ =& {180}^{\circ }\end{array}$

Therefore, $AG\text{//}BF$ (int. $\mathrm{\angle }$s supp.).

II is true.

In $\mathrm{\Delta }BCD$ and $\mathrm{\Delta }EFG$,

$\begin{array}{rcll}BC& =& EF& \text{(definition of regular polygon)}\\ CD& =& FG& \text{(definition of regular polygon)}\\ \mathrm{\angle }BCD& =& \mathrm{\angle }EFG& \text{(definition of regular polygon)}\end{array}$

$\therefore \mathrm{\Delta }BCD\cong \mathrm{\Delta }EFG$ (S.A.S.).

Hence, $BD=EG$ (corr. sides, $\cong \mathrm{\Delta }$).

III is true.

$\begin{array}{rcl}\mathrm{\angle }CAG& =& \mathrm{\angle }BAH-\mathrm{\angle }BAC-\mathrm{\angle }GAH\\ \mathrm{\angle }CAG& =& {135}^{\circ }-{22.5}^{\circ }-{22.5}^{\circ }\\ \mathrm{\angle }CAG& =& {90}^{c}irc\end{array}$

$\begin{array}{rcl}\mathrm{\angle }BDH& =& {\displaystyle \frac{1}{2}}\times \mathrm{\angle }CDE-\mathrm{\angle }BDC\\ \mathrm{\angle }BDH& =& {\displaystyle \frac{1}{2}}\times {135}^{\circ }-{22.5}^{\circ }\\ \mathrm{\angle }BDH& =& {45}^{\circ }\end{array}$

Hence, $\mathrm{\angle }CAG=2\mathrm{\angle }BDH$.