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Good afternoon to everybody. Today I was reading a chapter on an Euclidean geometry book related to the theorem of Ptolemys and the theorem of Brahmagupta and there was an exercise about how to construct a quadrilateral(not necessary inscribable) if we are given the length of its diagonals, the angle between them and also the two opposite angles of the quadrilateral. If for example we denote by ABCD the quadrilateral, and the angles ABC and ADC are given, then we know that the vertices of these two angles lie in the arc of two circles which intersect at the common chord AC. But from there I do not know how to construct the other diagonal which has a given angle with the first one (and a given length). Any ideas would be really helpful (also I do not know if there is any relation to the theorems of Ptolemys and Brahmagupta,I do not see any relation for now)

Petros Karajan
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  • So, given a quadrilateral $ABCD$ whose diagonals $AC$ and $BD$ intersect at $M$, you know the lengths $AC$ and $BD$ and you know the angles $\angle AMB,$ $\angle ABC,$ and $\angle ADC.$ Do you know any other angles? Because I see cases where the solution can be brought down to two possible quadrilaterals with this information, but both quadrilaterals satisfy the conditions. – David K Jan 08 '23 at 18:31
  • Yes, these are the given data. No , there are no other known angles. Maybe there is not unique solution though. How did you construct them? – Petros Karajan Jan 08 '23 at 18:35

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As you have already found out, the two endpoints of the diagonal $BD$ must lie one on each of two arcs from $A$ to $C,$ and these arcs can be constructed knowing the angles $\angle ABC$ and $\angle ADC.$

In particular, since one arc is the arc of points from which $AC$ subtends an angle $\angle ABC,$ you know that $B$ must be on that arc.

Supposing that we regard a quadrilateral and its mirror image as the same quadrilateral, it does not matter whether the angle $BD$ makes with $AC$ is "clockwise" or "counterclockwise", so we can assume WLOG that we can construct line segment $AE$ of the same length as $BD$ so that $AEDB$ is a parallelogram. Having done this, we also know that arc $ABC$ is on the opposite side of $AC$ from $E.$

Construct the parallelogram $AEFC.$ That is, $AE,$ $BD,$ and $CF$ are all parallel to each other and oriented in the same direction.

Now just as we know that $B$ lies on the arc $ABC,$ which is an arc we can construct, we know $D$ is somewhere on an arc $EDF,$ which is congruent to $ABC$ but translated to lie between $E$ and $F$ instead.

Having constructed arc $EDF,$ consider arc $ADC,$ which we can also construct from the information in the first paragraph and the knowledge that the arc must be on the opposite side of $AC$ from arc $ABC.$ The arcs $EDF$ and $ADC$ may intersect in no points (which tells us the given conditions of the quadrilateral are impossible), one point, or two points. In the case of one intersection there is a unique solution to the problem, but in the case of two intersections there are two solutions that in general will not be congruent, even allowing reflection.

I don't have any particular insights on how to apply the theorems of either Ptolemy or Brahmagupta here, I'm sorry to say.

David K
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  • Yees,thank you a lot David, that was really helpful ! Indeed I understand what you are saying. In fact, we can construct the parallelogram AEFC with AE=FC=BD(which has given length), and so that the angle between the line AE and AC is the same as the (given ) angle between the diagonals.. These are all constructible. . Now we construct an arc equal to the arc ABC(which is also constructible,since the angle at B is given) , but the new arc lies above the chord EF. – Petros Karajan Jan 08 '23 at 23:15
  • Now,if the new arc intersects the one which lies BELOW the chord AC(the one which the vertex D is supposed to lie on),call D the intersection point,then draw from A a parallel to ED, which intersects the above AC arc at a point B.This is the other point we are looking for. In fact,one checks easily that the triangles ABC and EDF are equal,hence AB is not only parallel to ED but also EQUAL. THus AEDB is parallelogram,hence AE=BD and angle between BD and AC is equal to the one between AE and AC. – Petros Karajan Jan 08 '23 at 23:36