$ABCD$ is an iscribable quadrilateral. Prove $AB+CD=AD+BC$.

Question

I am self-studying from Gerry Leversha's Crossing the Bridge, and I got stuck on this problem:

$ABCD$ is an iscribable quadrilateral. Prove $AB+CD=AD+BC$.

I attempted a proof using trigonometry, but I suspect there is a more elementary way.

My attempt is as follows: Let the centre be $O$ and join the vertices to $O$. Then define $\alpha:=\angle AOB,\beta:=\angle BOC,\gamma:=\angle COD,\delta:=\angle DOA$. Then $AB=2r\sin\alpha$ and similar results hold for the other sides. It suffices to prove $\sin\alpha+\sin\gamma=\sin\beta+\sin\delta$. I then tried to use the sum to product formulas and almost got there but not quite because $\cos\left(\frac{\alpha-\gamma}{2}\right)\neq\cos\left(\frac{\beta-\delta}{2}\right)$.

I also tried to use similar triangles, letting $E:=AC\cap BD$ and then it easily follows that $\triangle ABE\sim\triangle DCE$ and $\triangle BCE\sim\triangle ADE$, but then there are two different scale factors and the problem reduces to showing $AD/DC=(1+k)/(1+k')$, where $k,k'$ are the scale factors. I also could not show this.

Answer 1

I suppose inscribable means a circle can be inscribed in. If you mean this by statement, then we may write:

$AF=AB_1$

$FB=BE$

$CG=CE$

$GD=DB_1$

suming up theses relations we get:

$AF+FB+CG+GD=BE+CE+DB_1+AB_1$

Or:

$AB+CD=AD+BC$

This is the only possibility, in fact the correct statement for this result is:

A circle can be inscribed in the quadrilateral ABCD. Prove $AB+CD=AD+BC$.

Answer 2

The result as stated is false: consider an inscribed rectangle with width $>$ length, then $AB+CD\neq AD+BC$. As per Piquito's comment, the question should say circumscribed quadrilateral.