Prove that triangles $SBX,PCY,RAQ,ABC$ have the same area

Question

$ABC$ is a rt. angled triangle rt. angled at $A$. $ACPQ, BCYX, ABSR$ are squares drawn on $AC, BC , AB$ respectively. Prove that triangles $SBX,PCY,RAQ,ABC$ have the same area.

No clue except area of $ACY=1/2 .ACYL$ Please help.

Answer 1

Triangle $AQR$ is congruent to $ABC$.

Triangles $CPL$ and $BSJ$ in picture below are also congruent to $ABC$. But triangle $BXS$ has the same base $BX\cong BJ$ and the same altitude $SK$ as triangle $BSJ$, and triangle $PCY$ has the same base $CY\cong CL$ and the same altitude $PH$ as triangle $CPL$. Both $BXS$ and $PCY$ have then the same area as $ABC$.

Answer 2

denoting $$A_i$$ are the searched Areas then we have $$A_1=\frac{1}{2}ac\sin(180^{\circ}-\beta)$$ $$A_2=\frac{1}{2}ab\sin(180^{\circ}-\gamma)$$ $$A_3=\frac{1}{2}bc$$ $$A_4=\frac{1}{2}bc$$

Answer 3

If you rotate around $C$ for $-90^{\circ}$ a point $P$ to $P'$ ($Y$ goes to $B$) you see that $A,C,P'$ are colinear and that $C$ halves $AP'$. Also $\triangle CBP'\cong \triangle CYP$. So $BC$ is median for triangle $BAP'$ and thus $$S_{ABC} = S_{CBP'} = S_{CYP}$$

If you rotate around $B$ for $90^{\circ}$ you get with similar reasoning $S_{ABC} = S_{SBX}$.

Also it is obviously that $S_{ABC}= S_{ARQ}$ since $\triangle ABC\cong \triangle ARQ$.

Answer 4

Choose point T so that SBXT is a parallelogram. So $ST = BX = BC$. Also $\angle TSB = 180^\circ - \angle SBX = \angle ABC$. Can you prove that triangle SBT is congruent to ABC,and thus has the same area?

Both triangles SBX and SBT are "halves" of the parallelogram, so they also have the same area, and thus SBX has the same area as ABC.

Similarly for PCY.

RAQ is already congruent to ABC as it is.