Prove that two intersecting congruent chords of the same circle are divided by their intersection point into respectively congruent segments.
Source: Kiselev Geometry
The claim follows immediately from the Power of a Point Theorem; however, that theorem (or the Intersecting Chords Theorem) is not yet available.
My proof is below, to which I request verification and critique. I also would appreciate any insight into the remark at the end.
Let $AB, CD$ be congruent chords of the Circle with Center at $O$ intersecting at $X$. Assume WLOG that $AX \leq XB, CX \leq XD$. We will show that $AX \cong CX$ and $XB \cong XD$.
Drop the perpendicular from $O$ to $AB$ as $OP$ and to $CD$ as $OQ$. Since $AB \cong CD$, we have $OP \cong OQ$, as congruent chords are equidistant from the center. Then $\triangle OBP \cong \triangle ODQ$ by Hypotenuse-Leg. Likewise, $\triangle OXP \cong \triangle OXQ$ by Hypotenuse-Leg, completing the proof.
Remark: An alternate approach is that since circles are rotationaly symmetric, two congruent chords must be indistinguishable modulo rotation, and the claim follows immediately. For the same reason $XO$ must be a diameter and therefore a line of symmetry, again making the claim immediate. While I believe this is true, I wasn't able to turn this to a formal argument.
Update
I believe the symmetry approach is insufficient: Symmetry immediately shows the existence of a chord congruent to $AB$ that intersects $AB$ at $X$ with the claimed property, but it can't show its uniqueness: That is, perhaps there is a second chord congruent to $AB$, intersecting it at $X$, that does not have the claimed property.
Proving uniqueness is tougher; see Prove: One and only one congruent chord intersects a given chord at a given point.
DISCUSSION :
By & large , OP Proof is right.
Some "minor/finer" Points to consider :
"Let $AB, CD$ be congruent chords of the Circle with Center at $O$ intersecting at $X$" : I will be better to state that congruent chords are Equi-Distant from the Center. More-over , Proof uses that fact , hence high-lighting that is useful.
"Assume WLOG that $AX \leq XB, CX \leq XD$" : It is better to Prove this. More-over , we have to show that $P$ is between $X$ & $B$ and $Q$ is between $X$ & $D$ , which will then make it $BP+PX=BX$ and $DQ+QX=DX$
Only then we can claim the Congruence. Otherwise , it might occur that $BP+PX=BX$ and $DQ-QX=DX$ & Congruence will not work out.
[[ In other words , both segments have to be added to get overall segment. If we have to add in one case & subtract in other case , Congruence is lost ]]
"Remark: An alternate approach is that since circles are rotationaly symmetric, two congruent chords must be indistinguishable modulo rotation, and the claim follows immediately" : You have to explain what "modulo rotation" is !
"For the same reason $XO$ must be a diameter and therefore a line of symmetry, again making the claim immediate" : Every line though $O$ is automatically a Diameter , not necessary to use "modulo rotation" reason here.
Alternate Proof :
It is Essentially Equivalent what OP gave , because the configuration is very simple.
Congruent Chords are Equal , say $Z$ , in length.
Mid-Points $Q$ & $P$ will make it $Z/2$.
The tangents to Circle with Center at $O$ & radius $OP$ & $OQ$ will intersect at $X$.
Hence $QX=PX$
$AX$ & $CX$ are the left-over segments $Z/2-QX$ & $Z/2-PX$ , hence Equal & Congruent.