Let ABC be a triangle with angle B = 90^{circ | vedclass

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(D) In $	riangle ABC$,$\angle B = 90^{\circ}$. $AD$ is the angle bisector of $\angle A$. Draw a perpendicular from $D$ to $AC$,let it be $DE$. Since

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$\frac{10}{3}$

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(D) In $	riangle ABC$,$\angle B = 90^{\circ}$. $AD$ is the angle bisector of $\angle A$. Draw a perpendicular from $D$ to $AC$,let it be $DE$. Since $AD$ is the angle bisector,$DE = DB$ (distance from bisector to sides). Area of $	riangle ADC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AC 	imes DE = 10$. Given $AC = 6 	ext{ cm}$,we have $\frac{1}{2} 	imes 6 	imes DE = 10$. $3 	imes DE = 10 \implies DE = \frac{10}{3} 	ext{ cm}$. Since $DB = DE$,the length of $BD$ is $\frac{10}{3} 	ext{ cm}$.

Let $ABC$ be a triangle with $\angle B = 90^{\circ}$. Let $AD$ be the bisector of $\angle A$ with $D$ on $BC$. Suppose $AC = 6 \text{ cm}$ and the area of the $\triangle ADC$ is $10 \text{ cm}^2$. Then,the length of $BD$ in $\text{cm}$ is equal to

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