Condition for bisecting a diagonal – High School Geometry

Convex quadrilaterals here.

Consequently, the bisection condition to be established in today’s post applies only to convex quadrilaterals — quadrilaterals where none of the interior angles exceed $180^{\circ}$ .

Concave quadrilaterals also exist, but we won’t cover those.

Let $ABCD$

be a convex quadrilateral whose diagonals $AC$

and $BD$

intersect at $E$

. If $AC$

bisects $BD$

, PROVE that $\frac{AE}{EC}=\frac{d^2-a^2}{b^2-c^2}$

As usual, $a,b,c,d$ are the side-lengths $AB,BC,CD,DA$ . Assume for the time being that none of these sides are equal.

Let the length of diagonal $BD$ be $q$ . Since $AC$ bisects $BD$ , both $AE$ and $CE$ are medians in triangles $ABD$ and $BCD$ , respectively. Thus:

$AE^2=\frac{2a^2+2d^2-q^2}{4},~CE^2=\frac{2b^2+2c^2-q^2}{4}$

The angles $AEB$ and $CED$ are equal, so:

$\begin{align*} \cos(\angle AEB)&=\cos(\angle CED)\\ \frac{AE^2+BE^2-a^2}{2\times AE\times BE}&=\frac{CE^2+ED^2-c^2}{2\times CE\times ED}\\ \frac{AE^2+BE^2-a^2}{AE}&=\frac{CE^2+ED^2-c^2}{CE}\\ \frac{AE}{EC}&=\frac{AE^2+BE^2-a^2}{CE^2+ED^2-c^2}\\ &=\frac{\frac{2a^2+2d^2-q^2}{4}+\left(\frac{q}{2}\right)^2-a^2}{\frac{2b^2+2c^2-q^2}{4}+\left(\frac{q}{2}\right)^2-c^2}\\ &=\frac{d^2-a^2}{b^2-c^2} \end{align*}$

We later realized that applying Stewart’s theorem would have made the proof simpler. Same with the converse in example 2 below.

Let $ABCD$

be a convex quadrilateral whose diagonals $AC$

and $BD$

intersect at $E$

. If $\frac{AE}{EC}=\frac{d^2-a^2}{b^2-c^2}$

, PROVE that $AC$

bisects $BD$

To this end, consider the four triangles $AEB$ , $BEC$ , $AED$ , and $CED$ . For simplicity let’s set

$AE=p_1,~EC=p_2,~BE=q_1,~ED=q_2$

Thus the assumption $\frac{AE}{EC}=\frac{d^2-a^2}{b^2-c^2}$ can then be re-written as $\frac{p_1}{p_2}=\frac{d^2-a^2}{b^2-c^2}$ , or $p_1=\left(\frac{d^2-a^2}{b^2-c^2}\right)p_2$ . The angles $\angle AEB$ and $\angle BEC$ are supplementary, so:

$\begin{align*} \cos(\angle AEB)&=-\cos(\angle(BEC)\\ \implies\frac{AE^2+BE^2-AB^2}{2\times AE\times BE}&=-\left(\frac{BE^2+EC^2-BC^2}{2\times BE\times EC}\right)\\ \implies \frac{p_1^2+q_1^2-a^2}{2p_1q_1}&=-\left(\frac{q_1^2+p_2^2-b^2}{2q_1p_2}\right)\\ p_2(p_1^2+q_1^2-a^2)&=-p_1(q_1^2+p_2^2-b^2)\\ (p_1+p_2)q_1^2&=b^2p_1+a^2p_2-p_1p_2^2-p_1^2p_2\\ &=b^2\left[\left(\frac{d^2-a^2}{b^2-c^2}\right)p_2\right]+a^2p_2-p_1p_2(p_1+p_2)\\ &=\left(\frac{b^2d^2-a^2c^2}{b^2-c^2}\right)p_2-p_1p_2(p_1+p_2)\\ \therefore q_1^2&=\frac{p_2}{p_1+p_2}\left(\frac{b^2d^2-a^2c^2}{b^2-c^2}\right)-p_1p_2 \end{align*}$

Similarly, the angles $\angle CED$ and $\angle AED$ are supplementary; so:

$\begin{align*} \cos(\angle CED)&=-\cos(\angle AED)\\ \implies \frac{ED^2+EC^2-CD^2}{2\times ED\times EC}&=-\left(\frac{AE^2+ED^2-DA^2}{2\times AE\times ED}\right)\\ \frac{q_2^2+p_2^2-c^2}{2q_2p_2}&=-\left(\frac{p_1^2+q_2^2-d^2}{2p_1q_2}\right)\\ p_1(q_2^2+p_2^2-c^2)&=-p_2(p_1^2+q_2^2-d^2)\\ (p_1+p_2)q_2^2&=c^2p_1+d^2p_2-p_1p_2^2-p_2p_1^2\\ &=c^2\left[\left(\frac{d^2-a^2}{b^2-c^2}\right)p_2\right]+d^2p_2-p_1p_2(p_1+p_2)\\ &=\left(\frac{b^2d^2-a^2c^2}{b^2-c^2}\right)p_2-p_1p_2(p_1+p_2)\\ \therefore q_2^2&=\frac{p_2}{p_1+p_2}\left(\frac{b^2d^2-a^2c^2}{b^2-c^2}\right)-p_1p_2 \end{align*}$

The preceding calculation shows that $q_1^2=q_2^2$ , or $q_1=q_2$ . Thus $BE=ED$ and the diagonal $BD$ is bisected as desired.

Let $ABCD$

be a convex quadrilateral whose diagonals intersect at $E$

. Explain how the condition $\frac{AE}{EC}=\frac{d^2-a^2}{b^2-c^2}$

works in the case of parallelograms.

For parallelograms, opposite sides are equal, so we can set $b=d$ and $a=c$ and obtain $\frac{AE}{EC}=1$ . For squares and rhombuses, we need to take care: pretend that $a$ and $d$ are different, and that $b$ and $c$ are also different; at the same time, use the fact that opposite sides are equal. Doing this gives $\frac{AE}{EC}=1$ .

Let $ABCD$

be a convex quadrilateral whose diagonals intersect at $E$

. Explain how the condition $\frac{AE}{EC}=\frac{d^2-a^2}{b^2-c^2}$

works in the case of kites.

This is a special case because $a=d$ and $b=c$ and opposite sides are not equal in general kites. So we’re confronted with the indeterminate form $\frac{0}{0}$ . However, the longer diagonal in a kite can be as long as possible, and so the ratio $\frac{AE}{EC}$ evaluates differently depending on the kite in question. Does this somehow explain why $\frac{0}{0}$ is said to be undefined?

If quadrilateral $ABCD$

is such that diagonal $AC$

bisects diagonal $BD$

and $b=c$

, PROVE that $ABCD$

is a kite.

Let $E$ be the point of intersection of $AC$ and $BD$ . Since $b=c$ , triangle $BCD$ is isosceles, and so $CE$ is an altitude. Thus $AE$ is also an altitude. Since $BE=ED$ , this forces $a=d$ . So we obtain a kite.

Takeaway

Let $ABCD$ be a convex cyclic quadrilateral whose diagonals $AC$ and $BD$ intersect at $E$ . If the side-lengths are $AB=a$ , $BC=b$ , $CD=c$ , and $DA=d$ , then the following statements are equivalent:

$ab=cd$
$\frac{AE}{EC}=\frac{d^2-a^2}{b^2-c^2}$
$\frac{AE}{EC}=\frac{d^2+a^2}{b^2+c^2}$
$AC$ bisects $BD$ .

Notice the second and third statements.

Task

(Bisection consequence) Let $ABCD$ be a convex quadrilateral with side-lengths $AB=a$ , $BC=b$ , $CD=c$ , and $DA=d$ . If $AC$ bisects $BD$ and $ab=cd$ (quasi-harmonic), PROVE that $ABCD$ is cyclic.
(Basic characteristics) Let be a convex cyclic quadrilateral whose diagonals and intersect at . If (quasi-harmonic), PROVE that:
1. $p^2=AE^2+BE^2+CE^2+DE^2$
2. $pq\geq 2ab$ .