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The question #To prove two angles are equal when some angles are supplementary in a parallelogram has been solved. In the process of solving it, I found it is not that easy to draw the corresponding diagram..

Let’s start from P. Through it, 4 distinct, non-collinear rays PQ, PR, PS and PT are drawn with $\angle QPR$ and $\angle SPT$ are supplementary. A and B are pre-sectected points on PQ and PR respectively such that AB is of fixed length.

We are then supposed to find C and D on PR and PT respectively such that ABCD is parallelogram. [I don’t think the translation of a line is an acceptable Euclidean construction.]

Two questions:-

1) Can we prove that there always exist (at least one or may be only one) such a parallelogram?

2) If yes, what are the construction steps?

Community
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Mick
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  • If you want to draw a parallelogram of the kind you diagrammed, just reverse your proof. In the proof, given that you had such a parallelogram, you used cyclic construction. Now, reversely, draw any circle and pick four different points on the circumference to construct a cyclic quadrilateral. Pick one diagonal out of two. Cut the quadrilateral along the diagonal. At the moment you have two triangles. Move one along the path of another diagonal until the vertices linked by it meet each other. – James Pak Jul 20 '15 at 18:06
  • @JamesPak Yes. I can use some of the proven result to assist me in drawing such diagram. The thing that interests me is I have a hard time in drawing the parallelogram starting from point P. – Mick Jul 21 '15 at 03:41

1 Answers1

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1) Such a parallelogram does not always exist.

enter image description here

In this diagram, $\angle QPR$ and $\angle SPT$ are both right angles, so they are obviously supplementary. $m\angle RPS=30°$ and $\overline{AB}$ is drawn so that $\triangle ABP$ is isosceles. We can see from the orientations of $\overline{AB}$ and $\angle SPT$ that no points $C$ on $\overrightarrow{PS}$ and $D$ on $\overrightarrow{PT}$ can be chosen such that $\overline{AB}\parallel\overline{CD}$.

2) If the orientations of $\overline{AB}$ and $\angle SPT$ permit it then we can construct the points $C$ and $D$. In this diagram I have chosen different points $A$ and $B$ to allow this.

enter image description here

Choose an arbitrary point on $\overrightarrow{PS}$: I chose $S$ for simplicity. Construct point $U$ on $\overrightarrow{PT}$ so that $\overline{SU}\parallel\overline{AB}$. Draw the circle with center $U$ and radius $AB$. Mark the point $V$ at the intersection of the circle with line $\overleftrightarrow{SU}$.

Construct point $C$ on ray $\overrightarrow{PS}$ so that $\overline{CV}\parallel\overline{PU}$. Construct point $D$ on $\overrightarrow{PU}$ so that $\overline{CD}\parallel\overline{AB}$.

Then quadrilateral $ABCD$ is the desired parallelogram. That can be seen by examining parallelogram $CVUD$ and noting that $\overline{UV}=\overline{AB}$.

Rory Daulton
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  • Your first diagram clearly is an evident counter-example. 2) Your second diagram is asking for some relaxation of the given constraints. It is just like if I change the condition to “A is fixed, and then locate B, C, D”. That will probably generate many possible candidates.
    • – Mick Jul 21 '15 at 03:55
  • @Mick: My answer to 2) was more the idea "if such a parallelogram exists, this is how to construct it." It is also true that if the conditions of the problem are slightly changed (keep the length of $AB$ fixed but not the points $A$ and $B$) then the parallelogram exists, but that was not my main point. – Rory Daulton Jul 21 '15 at 09:23