समकोण में $\triangle \mathrm{ABC}, \angle \mathrm{ABC}=90^{\circ}$; $\mathrm{BN}$ $\mathrm{AC}, \mathrm{AB}=6 \mathrm{~cm}, \mathrm{AC}=10 \mathrm{~cm}$ के लंबवत है। तब AN : NC is

$\mathrm{BC}=\sqrt{10^{2}-6^{2}}=\sqrt{100-36}=\sqrt{64}=8 \mathrm{~cm}$

Area of $\triangle \mathrm{ABC}$,

$=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB}=\frac{1}{2} \times 8 \times 6=24 \text { sq.cm$

फिर से,

$\frac{1}{2} \mathrm{AC} \times \mathrm{BN}=24$

$\Rightarrow \frac{1}{2} \times 10 \times \mathrm{BN}=24 \Rightarrow \mathrm{BN}=\frac{24}{5}$

$\therefore \quad \mathrm{NC}=\sqrt{\mathrm{BC}^{2}-\mathrm{BN}^{2}}$

$\quad=\sqrt{64-\frac{576}{25}}=\frac{32}{5} \mathrm{~cm}$

$\quad \mathrm{AN}=10-\frac{32}{5}=\frac{50-32}{5}=\frac{18}{5}$

$\therefore \quad \mathrm{AN}: \mathrm{NC}=\frac{18}{5}: \frac{32}{5}=9: 16$