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I need to construct a rhombus $ABCD$ given that a line perpendicular to side $BC$ intersects diagonal $AC$ at point $M$ and its diagonal $BD$ at point $N$. The ratios $AM:MC = m:n$ and $BN:ND = p:q$ are given.

My attempt is as follows:

  1. Construct two arbitrary perpendicular lines $a$ and $b$. Label their intersection point as $P$.
  2. Choose point $B$ on line $a$ that is distinct from $P$.
  3. Choose point $N$ on line $b$ that is distinct from $P$.
  4. Contruct ray $BN$.
  5. On ray $BN$, construct a segment $ND = \frac{q}{p}\cdot BN$ in a way that point $N$ will belong to segment $BD$. We now have diagonal $BD$ matching the given condition.
  6. Label the midpoint of $BD$ as $O$.
  7. Through point $O$ construct a line perpendicular to $BD$. Label its intersection point with line $a$ as $C$.
  8. Construct a circle with center at $O$ of radius $OC$. Label its intersection point with line $CO$ distinct from $C$ as $A$.
  9. Connect the points $A$ and $B$, $B$ and $C$, $C$ and $D$, as well as $A$ and $D$. We have got rhombus $ABCD$ as its diagonals are perpendicular bisectors of one another.
  10. Label point of intersection of line $b$ with line $AC$ as $M$.

This construction will only give a correct ratio of $AM$ to $MC$ in one case. It's evident because there is only one point that divides segment $AC$ in a given ratio $m:n$, and it does not necessarily lie on line $b$. Therefore, my approach to the construction seems to be off.

What is the correct approach, and why?

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  • $\begingroup$ Your construction depends on choosing of points $B$ and $N$ and we have no any contradiction. Now, you need to calculate values of $PB$ and $PN$ such that $AM:MC=m:n$. $\endgroup$ Commented Jun 7 at 16:15
  • $\begingroup$ @MichaelRozenberg On a second thought, is it possible to determine the ratio of $BP$ to $PC$? The choice of point $P$ on $BC$ so that the perpendicular line through $P$ divides diagonals in given ratios should be unique. $\endgroup$ Commented Jun 7 at 16:23
  • $\begingroup$ To me, it seems that we can choose either $B$ or $N$ distinct from $P$, but not both. Having the point $B$, it's necessary to construct the point $N$, and vice versa. $\endgroup$ Commented Jun 7 at 16:40
  • $\begingroup$ Say me please. Points $M$ and $N$ on the line $a$ are they given? $\endgroup$ Commented Jun 7 at 16:43
  • $\begingroup$ @MichaelRozenberg As per problem's statement, a line perpendicular to side $BC$ intersects $AC$ and $BD$ at points $M$ and $N$, so yes. $\endgroup$ Commented Jun 7 at 16:44

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Note that triangle $OMN$ is similar to triangle $OBC$, where $O$ is the center of the rhombus. Hence: $$ {OB\over OC}={OM\over ON}={{m-n\over m+n}OC\over{p-q\over p+q}OB} $$ and you can compute from there the aspect ratio $OB/OC$: $$ {OB\over OC}=\sqrt{{m-n\over m+n}\cdot{p+q\over p-q}} $$

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  • $\begingroup$ Was writing the same thing. Then given the ratio and $MN$ as diameter, we construct $O$ $\endgroup$ Commented Jun 7 at 16:48
  • $\begingroup$ This was the key observation to nail the construction. Thanks for pointing this out! $\endgroup$ Commented Jun 8 at 14:46

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