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Prove that $\Box ABCD$ is a convex set whenever $\Box ABCD$ is a convex quadrilateral.

Things I know:

  • A set of points $S$ is said to be a convex set if for every pair of points $A$ and $B$ in $S$, the entire segment $AB$ is contained in $S$.

  • The diagonals of the quadrilateral $\Box ABCD$ are the segments $AC$ and $BD$

  • The angles of the quadrilateral $\Box ABCD$ are the angles $\angle ABC, \angle BCD, \angle CDA,$ and $\angle DAB$

  • A quadrilateral is said to be convex if each vertex of the quadrilateral is contained in the interior of the angle formed by the other three vertices.

How would I start this proof? Assuming that the quadrilateral $\Box ABCD$ is a convex quadrilateral.

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So I think we might be in the same geometry class. I think the best way to start this problem is to think of the two scenarios that could occur. Pick two points in the quadrilateral and think about what would happen if they were either the same point or two distinct points. Since you know you are working with a convex quadrilateral the line segment connecting the two points is also in ithe interior so the set must be convex. I'm not sure this will help but thats how I am thinking about it.

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Assume that $R=ABCD$ is convex quadrilateral Consider angle at $A$, $C_A$ :

$$ C_A=\bigg\{ u\bigg( s\overrightarrow{AB}+(1-s) \overrightarrow{AD}\bigg)| s\in [0,1],\ u\geq 0 \bigg\} $$

Since $C\in C_A$ so there is $u_0,\ s_0\in (0,1)$ s.t. $$ \overrightarrow{AC}=u_0\bigg( s_0\overrightarrow{AB}+(1-s_0) \overrightarrow{AD}\bigg) $$

Since $B,\ D$ is not in same side wrt line containing $[AC]$, then $$ R= \triangle ACB\cup \triangle ACD$$

And similar argument implies that $$R=\triangle BDA\cup \triangle BDC$$

Hence $\angle ABC,\ \angle BCD,\ \angle CDA,\ \angle DAB$ are strictly less than $\pi$ so that we complete the proof

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