Quasi harmonic quadrilaterals I – High School Geometry

Harmonic quadrilaterals are quadrilaterals that can be inscribed in a circle, with the additional property that the products of opposite sides are equal.

However, instead of requiring that the products of opposite sides are equal, we can stipulate that the products of consecutive sides are equal, and then call the resulting cyclic quadrilateral quasi harmonic.

Having this latter requirement (e.g. $ab=cd$ in the diagram below) yields some nice lateral properties for quadrilaterals that satisfy them.

Rectangles.

Any rectangle can be inscribed in a circle, and the products of consecutive sides are equal (e.g. $AB\times BC=CD\times DA$ below). Thus, rectangles are quasi-harmonic as per our definition. Note that a rectangle is not a harmonic quadrilateral, unless the rectangle is square.

Right kites and squares.

Right kites are both harmonic and quasi-harmonic. So are squares.

Consider quadrilateral $ABCD$

with vertices at $A(-7,7)$

, $B(-6,0)$

, $C(0,0)$

, and $D(2,4)$

. Verify that this is a quasi-harmonic quadrilateral.

We need to show that it’s cyclic, and that the products of the lengths of consecutive sides are equal.

side-lengths: $a=5\sqrt{2}$ , $b=6$ , $c=2\sqrt{5}$ , $d=3\sqrt{10}$
diagonals: $AC=7\sqrt{2}$ , $BD=4\sqrt{5}$
Ptolemy’s theorem: $AC\times BD=28\sqrt{10}$ , and $ac+bd=10\sqrt{10}+18\sqrt{10}=28\sqrt{10}$ . Confirms $ABCD$ is cyclic.
product of consecutive sides: $ab=30\sqrt{2}=cd$ . Quasi harmonic.

Consider quadrilateral $ABCD$

with vertices at $A(1,4)$

, $B(0,0)$

, $C(6,-6)$

, and $D(3,6)$

. Verify that this is a quasi-harmonic quadrilateral.

side-lengths: $a=\sqrt{17}$ , $b=6\sqrt{2}$ , $c=3\sqrt{17}$ , $d=2\sqrt{2}$
diagonals: $AC=5\sqrt{5}$ , $BD=3\sqrt{5}$
Ptolemy’s theorem: $AC\times BD=75$ , and $ac+bd=51+24=75$ . Confirms $ABCD$ is cyclic.
product of consecutive sides: $ab=6\sqrt{34}=cd$ . Quasi harmonic.

Consider quadrilateral $ABCD$

with vertices at $A\left(-\frac{12}{5},\frac{6}{5}\right)$

, $B(-3,6)$

, $C(1,8)$

, and $D(0,0)$

. Verify that this is a quasi-harmonic quadrilateral.

side-lengths: $AB=\frac{3}{5}\sqrt{65}$ , $BC=2\sqrt{5}$ , $CD=\sqrt{65}$ , $DA=\frac{6}{5}\sqrt{5}$
diagonals: $AC=\frac{17}{5}\sqrt{5}$ , $BD=3\sqrt{5}$
Ptolemy’s theorem: $AC\times BD=51$ , and $AB\times CD+BC\times DA=39+12=51$ . Confirms $ABCD$ is cyclic.
product of consecutive sides: $AB\times BC=6\sqrt{13}=CD\times DA$ . Quasi harmonic.

Takeaway

Let $ABCD$ be a convex cyclic quadrilateral with side-lengths $AB=a$ , $BC=b$ , $CD=c$ , and $DA=d$ . Then the following statements are equivalent:

$ab=cd$
diagonal $AC$ bisects diagonal $BD$ .

The longer diagonal in a quasi harmonic quadrilateral always bisects the shorter diagonal.

Task

(Bisection condition) Let be a convex cyclic quadrilateral with side-lengths , , , and . PROVE that:
1. diagonal $AC$ bisects diagonal $BD$ if and only if $ab=cd$
2. diagonal $BD$ bisects diagonal $AC$ if and only if $ad=bc$
3. diagonals $AC$ and $BD$ bisect each other if and only if $a=c$ and $b=d$ (rectangle).
(Basic characteristics) Let be a convex cyclic quadrilateral in which the side-lengths are , , , , and the diagonals are , . If (quasi-harmonic), PROVE that:
1. $2p^2=a^2+b^2+c^2+d^2$
2. $q^2=\frac{2(a^2+d^2)(b^2+c^2)}{a^2+b^2+c^2+d^2}$ .