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Quadrilaterals that has congruent opposite sides is parallelograms.

The following is a proof.

for quadrilateral ABCD,

AB = CD, BC = AD, AC = AC

hence ABC = CDA (SSS)

mBAC = mDCA (alternate interior angle theorem)

hence

AB // DC

mACB = mCAD (alternate interior angle theorem)

AD // BC

(Q.E.D)

but, I don't know why D, B is opposite by line AC for the alternate interior angle theorem.

How to prove it?

How to prove that ABCD is convex?

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  • $\begingroup$ The only quadrilaterals that aren't convex are ... I call them the Star Trek logo shape. You don't need to worry about it. To see why the other alt angle formula works, turn the diagram on its side, that might help your perception. $\endgroup$ – Matthew Daly Aug 16 at 12:18
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    $\begingroup$ This is a good question! It turns out that if you allow self-intersecting quadrilaterals, then $ABCD$ is not necessarily a parallelogram. $\endgroup$ – TonyK Aug 16 at 12:49
  • $\begingroup$ @TonyK I agree. Thank you for the link! $\endgroup$ – Michael Rozenberg Aug 17 at 5:55
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In your proof the point that $\Delta ABC\cong\Delta CDA$ is right, but the following step is wrong, because you need to prove that $B$ and $D$ are placed in the different sides respect to $AC$, which is not so obvious.

Indeed, let $B$ and $D$ are placed in the same side respect to $AC$.

Since $\Delta ABC\cong\Delta CDB$, we obtain: $$\measuredangle BAC=\measuredangle DCA$$ and $$\measuredangle BCA=\measuredangle DAC.$$

Now, if $\measuredangle BAC=\measuredangle DAC$ so rays $AB$ and $AD$ they are the same ray and rays $CB$ and $CD$ they are the same ray, which gives $B\equiv D$, which is a contradiction.

Let $\measuredangle BAC>\measuredangle DAC$.

Thus, the ray $AD$ is placed between sides of the angle $BAC$,

which says that the ray $AD$ intersects a side $BC$.

Let it happens in the point $K$.

By the same way we obtain that the ray $CB$ intersects a side $BC$ and let it happens in the point $L$.

Id est, lines $AD$ and $BC$ intersects in $\{K,L\}$, which says $K\equiv L$ and sides $BC$ and $AD$ have a common point, which is a contradiction again.

Thus, $B$ and $D$ are placed in the different sides respect to $AC$, which gives that $\angle BAC$ and $\angle DCA$ are alternate angles between $AB$ and $CD$ and secant $AC$, which says $AB||CD.$

By the same way we obtain $BC||AD$, which gives $ABCD$ is a parallelogram and we are done!

Actually.

This theorem is one of hardest theorems at the beginning of geometry.

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