Given $\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. If $2 m \angle P = 3 m \angle Q$ and $m \angle C = 100^{\circ}$,find $m \angle B$. (in $^{\circ}$)

In $\triangle PQR$,$PD \perp QR$ such that $D$ lies on $QR$. If $PQ = a$,$PR = b$,$QD = c$,and $DR = d$,prove that $(a+b)(a-b) = (c+d)(c-d)$.

In the given figure,if $\angle 1 = \angle 2$ and $\triangle NSQ \cong \triangle MTR,$ then prove that $\triangle PTS \sim \triangle PRQ.$

When a ladder leans on a wall to reach a height of $8 \, m$, its lower end rests $6 \, m$ away from the base of the wall. If the same ladder leans on the wall to reach a height of $6 \, m$, find the distance of its lower end from the base of the wall. (in $m$)

It is given that $\triangle DEF \sim \triangle RPQ.$ Is it true to say that $\angle D = \angle R$ and $\angle F = \angle P?$ Why?

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