Consequently, the bisection condition to be established in today’s post applies only to convex quadrilaterals — quadrilaterals where none of the interior angles exceed .
Concave quadrilaterals also exist, but we won’t cover those.
Check how many attacks we faced this week! Cause for concern?
![Rendered by QuickLaTeX.com ABCD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-563addbbd3bac1a4b850c391dabc7878_l3.png)
![Rendered by QuickLaTeX.com AC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-2509ee19f0ee692f93e2b93071b678f5_l3.png)
![Rendered by QuickLaTeX.com BD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-95329c4b4f11085d9454b642b291a0bb_l3.png)
![Rendered by QuickLaTeX.com E](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-ee84f23ec5accbe2212951a450e63b6d_l3.png)
![Rendered by QuickLaTeX.com AC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-2509ee19f0ee692f93e2b93071b678f5_l3.png)
![Rendered by QuickLaTeX.com BD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-95329c4b4f11085d9454b642b291a0bb_l3.png)
![Rendered by QuickLaTeX.com \frac{AE}{EC}=\frac{d^2-a^2}{b^2-c^2}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3c8c64d679e662f2ff854a5062e9e1ff_l3.png)
As usual, are the side-lengths
. Assume for the time being that none of these sides are equal.
Let the length of diagonal be
. Since
bisects
, both
and
are medians in triangles
and
, respectively. Thus:
The angles and
are equal, so:
We later realized that applying Stewart’s theorem would have made the proof simpler. Same with the converse in example 2 below.
![Rendered by QuickLaTeX.com ABCD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-563addbbd3bac1a4b850c391dabc7878_l3.png)
![Rendered by QuickLaTeX.com AC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-2509ee19f0ee692f93e2b93071b678f5_l3.png)
![Rendered by QuickLaTeX.com BD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-95329c4b4f11085d9454b642b291a0bb_l3.png)
![Rendered by QuickLaTeX.com E](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-ee84f23ec5accbe2212951a450e63b6d_l3.png)
![Rendered by QuickLaTeX.com \frac{AE}{EC}=\frac{d^2-a^2}{b^2-c^2}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3c8c64d679e662f2ff854a5062e9e1ff_l3.png)
![Rendered by QuickLaTeX.com AC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-2509ee19f0ee692f93e2b93071b678f5_l3.png)
![Rendered by QuickLaTeX.com BD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-95329c4b4f11085d9454b642b291a0bb_l3.png)
To this end, consider the four triangles ,
,
, and
. For simplicity let’s set
Thus the assumption can then be re-written as
, or
. The angles
and
are supplementary, so:
Similarly, the angles and
are supplementary; so:
The preceding calculation shows that , or
. Thus
and the diagonal
is bisected as desired.
![Rendered by QuickLaTeX.com ABCD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-563addbbd3bac1a4b850c391dabc7878_l3.png)
![Rendered by QuickLaTeX.com E](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-ee84f23ec5accbe2212951a450e63b6d_l3.png)
![Rendered by QuickLaTeX.com \frac{AE}{EC}=\frac{d^2-a^2}{b^2-c^2}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3c8c64d679e662f2ff854a5062e9e1ff_l3.png)
For parallelograms, opposite sides are equal, so we can set and
and obtain
. For squares and rhombuses, we need to take care: pretend that
and
are different, and that
and
are also different; at the same time, use the fact that opposite sides are equal. Doing this gives
.
![Rendered by QuickLaTeX.com ABCD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-563addbbd3bac1a4b850c391dabc7878_l3.png)
![Rendered by QuickLaTeX.com E](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-ee84f23ec5accbe2212951a450e63b6d_l3.png)
![Rendered by QuickLaTeX.com \frac{AE}{EC}=\frac{d^2-a^2}{b^2-c^2}](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-3c8c64d679e662f2ff854a5062e9e1ff_l3.png)
This is a special case because and
and opposite sides are not equal in general kites. So we’re confronted with the indeterminate form
. However, the longer diagonal in a kite can be as long as possible, and so the ratio
evaluates differently depending on the kite in question. Does this somehow explain why
is said to be undefined?
![Rendered by QuickLaTeX.com ABCD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-563addbbd3bac1a4b850c391dabc7878_l3.png)
![Rendered by QuickLaTeX.com AC](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-2509ee19f0ee692f93e2b93071b678f5_l3.png)
![Rendered by QuickLaTeX.com BD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-95329c4b4f11085d9454b642b291a0bb_l3.png)
![Rendered by QuickLaTeX.com b=c](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-0b523cca7cb248c9d5f7ef103ad6fdb6_l3.png)
![Rendered by QuickLaTeX.com ABCD](https://blog.fridaymath.com/wp-content/ql-cache/quicklatex.com-563addbbd3bac1a4b850c391dabc7878_l3.png)
Let be the point of intersection of
and
. Since
, triangle
is isosceles, and so
is an altitude. Thus
is also an altitude. Since
, this forces
. So we obtain a kite.
Takeaway
Let be a convex cyclic quadrilateral whose diagonals
and
intersect at
. If the side-lengths are
,
,
, and
, then the following statements are equivalent:
bisects
.
Notice the second and third statements.
Task
- (Bisection consequence) Let
be a convex quadrilateral with side-lengths
,
,
, and
. If
bisects
and
(quasi-harmonic), PROVE that
is cyclic.
- (Basic characteristics) Let
be a convex cyclic quadrilateral whose diagonals
and
intersect at
. If
(quasi-harmonic), PROVE that:
-
.