2018-II-19

Ans: D
I is true. In $\mathrm{\Delta }CDE$,

$\begin{array}{rcll}\mathrm{\angle }CDE& =& {\displaystyle \frac{(5-2)\times {180}^{\circ }}{5}}& \text{(}\mathrm{\angle }\text{ sum of polygon)}\\ \mathrm{\angle }CDE& =& {108}^{\circ }\end{array}$

Also,

$\begin{array}{rcll}CD& =& ED& \text{(regular pentagon)}\\ \mathrm{\angle }DCE& =& \mathrm{\angle }DEC& \text{(base }\mathrm{\angle }\text{s, isos. }\mathrm{\Delta }\text{)}\\ \mathrm{\angle }DCE& =& {\displaystyle \frac{1}{2}}\times ({180}^{\circ }–\mathrm{\angle }CDE)& \text{(}\mathrm{\angle }\text{ sum of }\mathrm{\Delta }\text{)}\\ \mathrm{\angle }DCE& =& {\displaystyle \frac{1}{2}}\times ({180}^{\circ }–{108}^{\circ })\\ \mathrm{\angle }DCE& =& {36}^{\circ }\end{array}$

Consider $\mathrm{\Delta }CDE$ and $\mathrm{\Delta }DEA$.

$\begin{array}{rcll}CD& =& DE& \text{(regular pentagon)}\\ DE& =& EA& \text{(regular pentagon)}\\ \mathrm{\angle }CDE& =& \mathrm{\angle }DEA& \text{(regular pentagon)}\end{array}$

Therefore, $\mathrm{\Delta }CDE\cong \mathrm{\Delta }DEA\text{(S.A.S.)}$.

Hence, $\mathrm{\angle }ADE=\mathrm{\angle }ECD={36}^{\circ }\text{(corr. }\mathrm{\angle }\text{s, }\cong \mathrm{\Delta }\text{s)}$.

Consider $\mathrm{\Delta }CDF$.

$\begin{array}{rcl}\mathrm{\angle }CDF& =& \mathrm{\angle }CDE–\mathrm{\angle }EDA\\ \mathrm{\angle }CDF& =& {108}^{\circ }–{36}^{\circ }\\ \mathrm{\angle }CDF& =& {72}^{\circ }\end{array}$

Also,

$\begin{array}{rcll}\mathrm{\angle }CFD& =& {180}^{\circ }–\mathrm{\angle }DCF–\mathrm{\angle }CDF& \text{(}\mathrm{\angle }\text{ sum of }\mathrm{\Delta }\text{)}\\ \mathrm{\angle }CFD& =& {180}^{\circ }–{36}^{\circ }–{72}^{\circ }\\ \mathrm{\angle }CFD& =& {72}^{\circ }\end{array}$

Since $\mathrm{\angle }CDF=\mathrm{\angle }CFD$, then $CF=CD\text{(sides opp. eq. }\mathrm{\angle }\text{s)}$.

II is true. In $\mathrm{\Delta }AFD$ and $\mathrm{\Delta }AFE$,

$\begin{array}{rcll}CD& =& AE& \text{(regular pentagon)}\\ \mathrm{\angle }DCF& =& \mathrm{\angle }EAF& \text{(corr. }\mathrm{\angle }\text{s, }\cong \mathrm{\Delta }\text{s)}\\ \mathrm{\angle }CFD& =& \mathrm{\angle }AFE& \text{(vert. opp. }\mathrm{\angle }\text{s)}\end{array}$

Therefore, $\mathrm{\Delta }AFD\cong \mathrm{\Delta }AFE\text{(A.A.S.)}$.

Hence, $CF=AF\text{(corr. sides, }\cong \mathrm{\Delta }\text{s)}$.

In $\mathrm{\Delta }ABF$ and $\mathrm{\Delta }CBF$,

$\begin{array}{rcll}AB& =& CB& \text{(regular pentagon)}\\ BF& =& BF& \text{(common side)}\\ AF& =& CF& \text{(proved)}\end{array}$

Therefore, $\mathrm{\Delta }ABF\cong \mathrm{\Delta }CBF\text{(S.S.S.)}$.

III is true. Since $\mathrm{\Delta }ABF\cong \mathrm{\Delta }CBF$, $\mathrm{\angle }BFC=\mathrm{\angle }BFA\text{corr. }\mathrm{\angle }\text{s, }\cong \mathrm{\Delta }\text{s)}$.

$\begin{array}{rcll}\mathrm{\angle }AFB& =& {\displaystyle \frac{1}{2}}({180}^{\circ }–\mathrm{\angle }AFE)& \text{(adj. }\mathrm{\angle }\text{s on a st. line)}\\ \mathrm{\angle }AFB& =& {\displaystyle \frac{1}{2}}({180}^{\circ }–{72}^{\circ })\\ \mathrm{\angle }AFB& =& {54}^{\circ }\end{array}$

Therefore, we have

$\begin{array}{rcl}\mathrm{\angle }AFB+\mathrm{\angle }EAF& =& {54}^{\circ }+{36}^{\circ }\\ & =& {90}^{\circ }\end{array}$