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

(N/A) Produce $CB$ to a point $D$ such that $BC = BD$ and join $AD$.
In $\Delta ABC$ and $\Delta ABD$,we have:
$BC = BD$ (By construction)
$AB = AB$ (Common side)
$\angle ABC = \angle ABD = 90^{\circ}$ (Given and by construction)
Therefore,$\Delta ABC \cong \Delta ABD$ by $SAS$ congruence rule.
So,$\angle CAB = \angle DAB$ by $CPCT$. Let $\angle BAC = x$,then $\angle CAB = x$ and $\angle DAB = x$.
Thus,$\angle CAD = \angle CAB + \angle BAD = x + x = 2x$.
Since $\angle BCA = 2 \angle BAC$,and $\angle BAC = x$,then $\angle BCA = 2x$.
In $\Delta ACD$,$\angle CAD = 2x$ and $\angle ACD = 2x$.
Since two angles are equal,the sides opposite to them are equal,so $AC = CD$.
Also,since $\Delta ABC \cong \Delta ABD$,$AC = AD$ by $CPCT$.
Thus,$\angle ADC = \angle ACD = 2x$.
Since all three angles of $\Delta ACD$ are $60^{\circ}$ (as $2x + 2x + 2x = 180^{\circ} \implies x = 30^{\circ}$),$\Delta ACD$ is an equilateral triangle.
Therefore,$AC = CD = AD$.
Since $CD = BC + BD$ and $BC = BD$,we have $CD = BC + BC = 2BC$.
Hence,$AC = 2BC$.