$A$ $15 \, m$ long bamboo leans on a wall to reach the height of $12 \, m$ on the wall. Then,the lower end of the bamboo is $\ldots \ldots \, m$ away from the base of the wall.

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $AB = 5.2$,$AC = 10.4$,and $BD = 3.8$,find $BC$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is an altitude to the hypotenuse $\overline{AC}$. Then the correspondence $ADB \leftrightarrow \ldots$ between $\Delta ADB$ and $\Delta BDC$ is a similarity.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Prove that $\frac{1}{BM^2} = \frac{1}{AB^2} + \frac{1}{BC^2}$.

In $\Delta ABC$ and $\Delta PQR$,$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$. If $AB = 15$,$PQ = 20$ and $BC = 12$,find $QR$.

In trapezium $ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{P\}$. If $PA = 10$,$PC = 15$ and $PD = 12$,then $BD = \ldots$