Bisecting a diagonal in a cyclic quad

Let $ABCD$

be a cyclic quadrilateral with side-lengths $AB=a$

, $BC=b$

, $CD=c$

, and $DA=d$

. Let $AC$

and $BD$

be the diagonals. Then $AC$

bisects $BD$

if and only if $ab=cd$

, and $BD$

bisects $AC$

if and only if $ad=bc$

Let $ABCD$

be a convex quadrilateral whose diagonals intersect at $E$

. If $ABCD$

is cyclic, PROVE that $\frac{AE}{EC}=\frac{ad}{bc}$

and $\frac{BE}{ED}=\frac{ab}{cd}$

Use similar triangles.

Let $ABCD$

be a convex quadrilateral whose diagonals intersect at $E$

. If $\frac{AE}{EC}=\frac{ad}{bc}$

and $\frac{BE}{ED}=\frac{ab}{cd}$

, PROVE that $ABCD$

is cyclic.

This is the converse of the previous statement in example 1. One proof of it uses similar triangles and the law of sines.

Let $ABCD$

be a convex cyclic quadrilateral whose diagonals intersect at $E$

. PROVE that diagonal $AC$

bisects diagonal $BD$

if, and only if, $ab=cd$

By example 1 we have that $\frac{BE}{ED}=\frac{ab}{cd}$ . If $AC$ bisects $BD$ , then $BE=ED$ . And so $ab=cd$ . Similarly for the converse.

Let $ABCD$

be a convex cyclic quadrilateral whose diagonals intersect at $E$

. PROVE that diagonal $BD$

bisects diagonal $AC$

if, and only if, $ad=bc$

By example 1 we have that $\frac{AE}{EC}=\frac{ad}{bc}$ . If $BD$ bisects $AC$ , then $AE=EC$ . And so $ad=bc$ . Similarly for the converse.

Among convex cyclic quadrilaterals, PROVE that only rectangles have the two diagonals bisecting each other.

This follows from example 3 and example 4 above, since $ab=cd$ and $ad=bc$ yield $a=c$ and $b=d$ .

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Takeaway

Let $ABCD$ be a convex cyclic quadrilateral. Then the following statements are equivalent:

$ABCD$ is a rectangle
diagonals $AC$ and $BD$ bisect each other.

Task

(Easy verification) Let be a convex cyclic quadrilateral with vertices at , , , . Verify that:
1. diagonal $AC$ bisects diagonal $BD$
2. $2AC^2=AB^2+BC^2+CD^2+DA^2$
3. $BD^2=\frac{2(AB^2+DA^2)(BC^2+CD^2)}{AB^2+BC^2+CD^2+DA^2}$ .