In $\triangle PQR$,$\angle R = \angle P$,$QR = 4 \, cm$,and $PR = 5 \, cm$. Then the length of $PQ$ is (in $cm$):

In the figure,$ABC$ is a right triangle,right-angled at $B$,such that $\angle BCA = 2 \angle BAC$. Show that the hypotenuse $AC = 2 BC$.

In $\Delta PQR$,$PQ = PR$ and $\angle R = 40^{\circ}$,then $\angle P = \dots$ (in $^{\circ}$)

Line segments $AB$ and $CD$ bisect each other at $P$. If $PA = PD$ and $PB = PC$,prove that $AC = BD$.

In $\Delta PQR$,$S$ is any point in its interior. Prove that $SQ + SR < PQ + PR$.

It is given that $\triangle ABC \cong \triangle FDE$. If $AB = 5 \, cm$,$\angle B = 40^{\circ}$,and $\angle A = 80^{\circ}$,then which of the following is true?