In the figure, $ABCD$ is a square. If MN is perpendicular to $PQ$ and $P$ is any point, calculate $x$.(Answer:$1u^2$)
$\angle ANM \cong \angle NMC\\ \angle NMB \cong \angle MND$
$MN$ is common side
$$AB=CD$$
Is this enough to say that the two quadrilaterals are congruent?
On
The right angles make the quad. PMCQ cyclic. Then, $\angle PMC= \angle PQD$. Therefore, quad. PMCQ and quad. PQDN are equi-angular.
In addition, 'the two quadrilaterals are equal in area' implies they are congruent.
Then, PM of quad. PMCQ = PQ of quad PQDN and PQ of quad. PMCQ = PN of quad. PQDN. Therefore, PM = PQ = PN. (*)
The same logic extends to the third.
The logic of (*) plus the fourth quadrilateral is equi-angular should be enough to show that those four quadrilaterals are equal in area.
On
We can and do assume that the side of the square is one. Let us fix the position of $MN$, as shown in the picture below, and denote by $a,b$ the lengths of the segments $ND$ an $MC$.
We are searching first for a position of the segment $PQ$, with $PQ\perp MN$, and
Since $[MNDC]$ is fixed, and the slide of a line $PQ$ "downwards", parallel to itself on a direction perpendicular to $MN$, is a continuous movement cutting more and more from $[PADQ]$, the position of $PQ$ is unique. For short, the area $[PADQ]$ is a decreasing continuous function, seen as a function of $R\in[MN]$, which is moving from $M$ to $N$, and determining $PQ$. We know this position, it is realized by a $90^\circ$ rotation of $MN$ in the direction that brings the square $ABCD$ into $BCDA$, and $M$ in $Q$, $N$ in $P$, $MC=b$ in $QD=b$, $DN=a$ in $AP=a$.
So far we have arranged only that $U=[ANRP]=[CMRQ]$. Now we use the information that these areas are equal to $[DQRN]$. Again, fix the trapezium $APQD$ in sight, and let $R$ slide on $PQ$ from $P$ to $Q$, then draq the perpendicular $RN$. It is clear that the area $[APRN]$ is a continuous, increasing function of $R$, so there exists a unique point $R$ realizing $[APRN]$ as the half of $[APQD]$. For this one point record the special value $AN:ND$. Now do the same with the trapezium $DNMC$, congruent with $APQD$. Consider again the point $R$ sliding on $NM$, it has by the same argument an existing unique position of $RQ$ to cut $[DNMC]$ in two equal areas, and the corresponding singular point $Q$ realizes the same proportion: $$ \frac{AN}{ND}= \frac{DQ}{QC}\ ,\qquad\text{ i.e. }\qquad \frac{1-a}a=\frac{b}{1-b}\ . $$ This gives $a=1-b$, so both $MN$ and $PQ$ are passing through the center of symmetry of the square. So the fourth piece $BPRM$ is congruent to the other three pieces, thus having the same area.
$\square$
On
Let $O=MN\cap PQ$. Draw line segments $M'N'$ and $P'Q'$ through $O$ such that $M'N'\perp BC$, $P'Q'\perp AB$, $ON'=a$, $OQ'=b$, $\angle QOQ'=\theta$ as in the figure and let $k$ be the side length of the square. Adding and substracting the appropriate triangles after the rotation, we have $$a(k-b)-\frac12(k-b)^2\tan\theta+\frac12a^2\tan\theta=1\tag1$$ $$ab-\frac12a^2\tan\theta+\frac12 b^2\tan\theta=1\tag2$$ $$b(k-a)-\frac12b^2\tan\theta+\frac12(k-a)^2\tan\theta=1\tag3$$ If $a=b$, then $a=b=1$, $k=2$ and we are done. Assume that $a\neq b$. From $(1),(2),(3)$ we can obtain $$\tan\theta=\frac{b-a}{a+b-k}=\frac{2-2ab}{b^2-a^2}=\frac{4-k(a+b)}{k(b-a)}.$$ It is not so difficult to see that $k$ must be $2$. But, then $a=b=1$ contradicting the assumption.
Let $P'\in\overline{CD}$ be chosen so that $\overline{PP'}$ is parallel to $\overline{AD}$, and let $N'\in\overline{BC}$ be chosen so that $\overline{NN'}$ is parallel to $\overline{CD}$. Because $\overline{PQ}$ and $\overline{MN}$ are perpendicular, we see that $\angle QPP'\cong\angle MNN'$ and so $\triangle PP'Q\cong \triangle NN'M$. Thus, $P'Q = N'M$.
The areas of trapezoids $ADQP$ and $NDCM$ are equal, and so, since they have equal heights, $AP + DQ = ND + CM$. This means that $2AP + P'Q = 2ND + N'M$, and so $AP=ND$. Similarly, the areas of trapezoids $BPQC$ and $BANM$ are equal, and so $QC=MB$. However, I cannot continue at this stage.