• Mar 3rd 2008, 12:57 PM
Ultros88
Hello all,
I'm having difficulty trying to prove that the quadrilateral formed by the bisectors of the angles of any quadrilateral has its opposite angles supplemental. I just can't for the life of me see how equate the necessary angles. I'm sure it's easy. Any help would be appreciated.

Thanks,
Ultros
• Mar 3rd 2008, 01:36 PM
Plato
Quote:

Originally Posted by Ultros88
I'm having difficulty trying to prove that the quadrilateral formed by the bisectors of the angles of any quadrilateral has its opposite angles supplemental.

Are you sure of that statement?
Consider a square. The bisectors of the angles are just the diagonals.
• Mar 3rd 2008, 02:46 PM
Ultros88
Clarification
Sorry, I forgot to mention that there are cases where it will not hold such as when the quadrilateral is a square or if it is concave (?). In any case I am going through the exercises in a book and this is one of the few that I have gotten stuck on. Help proving it, in the cases where it is true, would be appreciated.

Thanks for clearing up those cases, Plato. If it cannot be done at all... let me know
-Ultros
• Mar 3rd 2008, 04:34 PM
roy_zhang
Given quadrilateral ABCD, and AH, BE, CG and DF are bisectors of $\angle A, \angle B, \angle C$ and $\angle D$ respectively, the quadrilateral EGFH is formed by these angle bisectors.
By definition, we have $\angle 1+\angle 2+\angle3+\angle4 =\frac{1}{2}(360^{\circ})=180^{\circ}$;
Let's consider $\triangle ABE$, we have $\angle3+\angle4+\angle6=180^{\circ}$,
And consider $\triangle CDF$, we have $\angle1+\angle2+\angle5=180^{\circ}$;
Hence, $\angle1+\angle2+\angle3+\angle4+\angle5+\angle6=36 0^{\circ}$,
substitute $\angle 1+\angle 2+\angle3+\angle4 =180^{\circ}$, we got: $\angle5+\angle6=180^{\circ}$. This shows that the opposite angles $\angle5$ and $\angle 6$ are supplemental, since the sum of inner angles of a quadrilateral is $360^{\circ}$, we have $\angle7+\angle8=180^{\circ}$
What would happen if $m\left( {\angle 1} \right) = m\left( {\angle 4} \right)$?