Show that if AB+CD=AD+BC then the quadrilateral $ABCD$ is tangential

Question

Consider a convex quadrilateral $ABCD$ . Show that the quadrilateral $ABCD$ is tangential if and only:$AB+CD=AD+BC$

Answer 1

Theorem:We know that in any tangential quadrilateral $ABCD$ we have $AB+CD=AD+BC$

proof:

lemma:The tangents that we draw from a point to a circle is equal.

because two triangles are equal tangents are equal too.

Using this lemma we can proof this theorem like this picture.

Now we want to proof the opposite of the theorem:

Think that The theorem holds and $ABCD$ isn't tangential then follow this picture:

according to the picture $AB'CD$ is tangential then we have:

$c+AB'=d+CB'$also we have $a+c=b+d$ from if we subtract the second one from the first one we have:$BB'=b-CB'$ and then $BB'+CB'=b$.Using the inequality of triangle we can know it never happens because:

Because the equality is false our assumption is false too.Then we can know $ABCD$ is tangential.