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?
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.
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.
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 .
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?
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.
- (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: