Converse of the Power of a Point Theorem

Ok, now that we know Power of a Point, let's go on to the converse, i.e. if the five points $A, B, C, D, E$ satisfy $AC \cap BD = E$ and $AE \cdot EC = BE \cdot ED$, then in fact the quadrilateral $ABCD$ is cyclic.

To prove this, we repeat the similar triangles argument to prove Power of a Point, but do it in reverse. Also draw in the segments $AB$, $DC$ to complete the triangles $AEB$ and $DEC$.

Now, we can manipulate $AE \cdot EC = BE \cdot ED$ to get $\frac{AE}{EB} = \frac{ED}{EC}$. Also, we have that $\angle AEB = \angle DEC$ by vertical angles. The previous 2 statements imply that $\triangle AEB \sim \triangle DEC$ by SAS similarity.

Now, since the 2 triangles are similar, we can match similar parts to get that $\angle EAB = \angle EDC$. But then $\angle CAB = \angle EAB = \angle EDC = \angle BDC$, so $\angle CAB = \angle BDC$. Therefore, the quadrilateral $ABCD$ is cyclic.

This proof works in the case when $E$ is inside the quadrilateral $ABCD$, but it's not completely clear that the proof works the same way when $E$ is outside. Let's do that case to check that you understand the proof.

In the above diagram, $EB \cdot EA = EC \cdot ED$, and we wish to show that $ABCD$ is cyclic.

In this case actually, we don't even need vertical angles. The 2 triangles that we're going to look at here actually will share the angle $\angle BEC$.

Now we can finish the proof. The above similarity gives that $\angle EAD = \angle ECB = 180-\angle DCB$. But $\angle EAD = \angle BAD$, so therefore, $\angle BAD = 180 - \angle DCB \implies \angle BAD + \angle DCB = 180$, so since opposite angles sum to $180^\circ$, we once again get that $ABCD$ is cyclic. Note that in this case, we used that opposite angles sum to $180^\circ$, instead of inscribed angles being equal (though if you learn directed angles, these are both actually equivalent).