A special cyclic quadrilateral II – High School Geometry

Following our previous discussion, we now consider quadrilaterals:

where a pair of interior opposite angles are both $90^{\circ}$
with one side and one diagonal having equal lengths.

For today’s purpose, let’s give the pseudonym special to cyclic quadrilaterals having the above two properties, and then prove that in any such special quadrilateral:

there’s a side whose length is twice the distance between the midpoints of the diagonals
the diagonals $p,q$ and the distance $x$ between the midpoints of the diagonals are related via
(1) $\begin{equation*}p^2=q^2+4x^2\end{equation*}$

Required knowledge

Euler’s quadrilateral formula is the main background needed in order to follow today’s discussion. It’s basically the equation

(2) $\begin{equation*}a^2+b^2+c^2+d^2=p^2+q^2+4x^2 \end{equation*}$

connecting the side-lengths $a,b,c,d$ of a quadrilateral with the diagonals $p,q$ and the distance $x$ between the midpoints of the diagonals.

Right kites

In a right kite, equation (2) simplifies to equation (1), that is $p^2=q^2+4x^2$ , as shown in our first example below.

Let $ABCD$

be a right kite with longer diagonal of length $p$

and shorter diagonal of length $q$

. If $x$

is the distance between the midpoints of the diagonals, PROVE that $\left(\frac{p}{2}\right)^2=\left(\frac{q}{2}\right)^2+x^2$

Consider the right kite shown below, with longer diagonal $AC=p$ , and shorter diagonal $BD=q$ .

Since $d=a, c=b$ in the kite shown above and $p^2=a^2+b^2$ because it’s a right kite, Euler’s quadrilateral formula (2) gives

$\begin{equation*} \begin{split} a^2+b^2+b^2+a^2&=p^2+q^2+4x^2\\ 2p^2&=p^2+q^2+4x^2\\ \therefore p^2&=q^2+4x^2 \end{split} \end{equation*}$

Let a cyclic quadrilateral $ABCD$

with side-lengths $a,b,c,d$

be special, in the sense of the two properties at the introduction. PROVE that $p^2=q^2+4x^2$

This example shows that apart from right kites, there are other (cyclic) quadrilaterals that satisfy equation (1), including the ones with the pseudonym special, which is our focus for today.

Such a special quadrilateral $ABCD$ contains two interior opposite angles measuring $90^{\circ}$ ; place them at $B$ and $D$ .

If we let diagonal $AC=p$ , then $p^2=a^2+b^2=c^2+d^2$ , and so by Euler’s quadrilateral formula (2):

$2p^2=p^2+q^2+4x^2\implies p^2=q^2+4x^2$

Let a cyclic quadrilateral $ABCD$

with side-lengths $a,b,c,d$

be special, in the sense of the two properties at the introduction. PROVE that there is a side whose length is twice the distance between the midpoints of the diagonals.

Using the same diagram/notation for the previous example, we have:

$c^2+d^2=p^2=q^2+4x^2$

Since one of the diagonals has the same length as one of the sides, it can’t be diagonal $AC$ because $AC$ is the diameter. So let’s suppose diagonal $BD=q$ as the same length as side $DA=d$ . Then the above equation gives $c=2x$ .

Rare kind

Consider a right kite $ABCD$

in which the interior angles at $A$

and $C$

are $120^{\circ}$

and $60^{\circ}$

. Show that $ABCD$

is special in today’s terms.

Diagonal $BD$ will now have the same length as sides $BC$ and $CD$ .

Show that the right kite in example 4 can be obtained from any $30^{\circ}-30^{\circ}-120^{\circ}$

isosceles triangle $ABC$

Suppose that $\angle B=120^{\circ}$ . Let $O$ and $H$ be the circumcenter and orthocenter of the parent triangle $ABC$ . Then the quadrilateral $OAHC$ is a particularly pleasant right kite.

Takeaway

Let $ABCD$ be a quadrilateral with side-lengths $a,b,c,d$ , diagonals $AC=p,BD=q$ , and $x$ the distance between the midpoints of the diagonals. If the interior angles at $B$ and $D$ are both $90^{\circ}$ , then the following statements are equivalent:

$c=2x$
$q=d$ .

Task

(Special property) Let be special, as per the definition at the introduction; specifically, let the interior opposite angles at and be both , and let diagonal have the same length as side . PROVE that:
1. the radius through $D$ is parallel to side $BC$
2. the orthocenter of triangle $BCD$ is a reflection of vertex $D$ over side $BC$ .