1. Oct 29, 2008

### mathstudent88

Here is the problem:

If the diagonals of a quadrilateral intersect each other, then the quadrilateral is convex.

Proof:

Let ABCD be a convex quadrilateral. Since quadrilateral ABCD is convex, A and D are on the same side of line BC, and D and C are on the same side of line AB. Thus D is a member of the int(angle ABC). With the Crossbad theorem, BD intersect AC = {P} where C-P-R. So AC intersect BC = {Q} where D-Q-R. Since A, B, C, D are noncollinear points P=Q. So AC intersect BD = {P} = {Q}. Which proves that AC intersects PR = the empty set. Since a convex quadrilateral has the property that its diagonals intersect then ABCD is conves.

How is this? I really didn't know what to do for it. Can someone please help me with it?

Thank you!

2. Oct 29, 2008

### HallsofIvy

Staff Emeritus
No. You cannot start with "Let ABCD be a convex quadrilateral". That is what you are asked to prove. Start with the hypotheis that the diagonals intersect. Or you can use "indirect" proof: Suppose the quadrilateral is NOT convex. Then what can you say about the diagonals?

[/quote]How is this? I really didn't know what to do for it. Can someone please help me with it?

Thank you![/QUOTE]

3. Oct 30, 2008

### mathstudent88

Is this better?

If A and B are on the same side of the line CD and B and C are on the same side of the line DA then B is in the interior of angle ADC. The Crossbar theorem says that the ray DB intersects the segment AC at some point P.
Similarly A is in the interior of angle BCD which implies ray CA intersects segment BD at some point Q.
Thus P and Q lie both on the lines AC and BD.
Being diagonals, these lines are not the same and so P=Q is the common point of intersection.
Because of this, these four condition hold:
A and B are on the same side of line CD
B and C are on the same side of line DA
C and D are on the same side of line AB
D and A are on the same side of line BC