Problem: Construct a parallelogram, given (the size of) its diagonals $D_1, D_2$ and altitude $A$. Source: Kiselev's Geometry.
My solution is below. I request verification, suggestions for improvement, or alternate approaches.
Construction:
Draw two parallel lines $\ell, m$ with distance $A$. Draw the perpendicular between them and bisect it at $X$.
Draw circle $C_1$ with center $X$ and radius $\frac {D_1} 2$; it will intersect $\ell$ at two points and $m$ at two points. Label the left intersection of $C_1$ and $\ell$ as point $P$, and the right intersection of $C_1$ and $m$ as $R$.
Similarly, draw circle $C_2$ with center $X$ and radius $\frac {D_2} 2$; it will intersect $\ell$ at two points and $m$ at two points. Label the right intersection of $C_2$ and $\ell$ as point $Q$, and the right intersection of $C_2$ and $m$ as $S$.
$PQRS$ is the desired parallelogram.
Proof: The diagonals of any parallelogram bisect each other at a point called the center. Consequently, the perpendicular from one base to the center is colinear and congruent to the perpendicular from the opposite base to the center. This means that an altitude drawn through the center will have its midpoint at the center; thus, $X$ is the center of the parallelogram.
The rest of the proof follows trivially.
Questions:
- Is my construction and proof correct?
- Did I omit any details that need support? I feel justified in concluding "The rest of the proof follows trivially" but would request if you agree.
- How could the exposition be improved?
- Is there an alternate approach?
Another solution:
$1.$ Draw the length of altitude $A$.
$2.$ Draw two parallel lines passing through the endpoints of altitude(say one of them is P) which are perpendicular to altitude.
$3.$ Let $D_1=PR$. Draw a circle centered at $P$ with radius $PR$. Then $R$ is the another vertex(opposite of $P$) of parallelogram on the opposite side.
$4.$ Determine midpoint of $PR$, say $M$.
$5.$ Draw a circle centered at M with radius $\frac{D_2}{2}$. Let this circle tangent the parallel lines at $Q$ and $S$.
$6.$ Then $PQRS$ is the parallelogram.