In parallelogram $ABCD$,$AB^{2} + BC^{2} = 260$ and $AC = 18$. Then $BD = \ldots$

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{AD}$ is a median. Prove that $AC^{2} = AD^{2} + 3BD^{2}$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BE}$ is a median. If $AB = 15$ and $BE = 8.5$,find $BC$.

In $\Delta ABC$,$AB = AC$ and $m \angle A = 90^\circ$. If $BC = \sqrt{2} a$,then the area of $\Delta ABC$ is $\ldots \ldots \ldots \ldots$ $(a \in R^+)$.

In $\Delta XYZ$,the bisector of $\angle X$ intersects $\overline{YZ}$ at $M$. If $XY = 3.6$,$XZ = 4.2$ and $MZ = 2.8$,find $YM$.

Prove that if in a triangle the square of one side is equal to the sum of the squares of the other two sides,then the angle opposite the first side is a right angle.