**(a)**
(i) In ΔDBA and ΔDAC: $\angle BDA=\angle ADC=90°$ (each). Since $\angle A=90°$, $\angle DAB+\angle DAC=90°$; also in ΔDBA, $\angle DBA+\angle DAB=90°$. So $\angle DBA=\angle DAC$. By AA, $\triangle DBA\sim\triangle DAC$.
(ii) From similarity: $\frac{DB}{DA}=\frac{DA}{DC} \Rightarrow DA^2=DB\times DC$
(iii) $DA^2=9\times16=144 \Rightarrow DA=12$ cm. $$\text{Area of }\triangle ABC=\frac12\times BC\times AD=\frac12\times(9+16)\times12=\frac12\times25\times12=150\ cm^2$$

**(b)** (Basic Proportionality Theorem) **Given:** ΔABC, DE||BC, D on AB, E on AC. **To Prove:** $\frac{AD}{DB}=\frac{AE}{EC}$. **Construction:** Join BE,CD; draw $DM\perp AC$, $EN\perp AB$.
$$\frac{ar(\triangle ADE)}{ar(\triangle DBE)}=\frac{AD}{DB},\quad\frac{ar(\triangle ADE)}{ar(\triangle DEC)}=\frac{AE}{EC}$$
Since $ar(\triangle DBE)=ar(\triangle DEC)$ (same base DE, between same parallels), $$\frac{AD}{DB}=\frac{AE}{EC}$$ Hence proved.