Problem Statement:
Suppose the straight line $AB$ of $\triangle ABC$ is bisected at $C$ and the perpendiculars $AX$, $BY$, $CZ$ are drawn to any straight line $OP.$ Prove that:
i) if $A$, $B$ are on the same side of $OP,$ then $2CZ = AX + BY.$
ii) if $A$, $B$ are on the opposite sides of $OP,$ then $2CZ =$difference of $AX$ and $BY$.
[from pg. 74, question number 21. in Challenge and Thrill of Pre-College Mathematics]
How can $C$ be a vertex of $\triangle ABC$ and
also where the side $AB$ of the triangle is bisected?
I think it might be a typo in the question. Assuming that they meant $O$ instead of $C$ as the bisector of $AB$, is there any case when the points $A$ and $B$ lie on the same side of a line drawn through their bisector (as specified in (i))?
I have searched elsewhere and in my books as well without much luck, so any help would be appreciated.
Edit:
Oops, (ii) should be 'opposite sides' not 'same side'. Also yeah, I find the wording confusing. In the degenerate case, if I take $OP \parallel AB$ both $A$ and $B$ lie on the same side of $OP$ then the perpendiculars $AX$, $BY$, $CZ$ have the same length so $2CZ = AX + BY$. So that's case (i).
For case (ii), if I take $OP$ through $C$ then $CZ$ = 0 and then if I prove that $\triangle ACX \cong \triangle BCY$ I can say that $AX = BY \implies 2CZ = AX – BY = 0.$
But I am not sure how to go about proving (ii) in general for 'any straight line $OP.$'
Best Answer
As @dxiv pointed out in the comment, the statement means $C$ is midpoint of $AB$. In that case part i) becomes trivial.
For part ii) when $OP$ does not pass through $C$ we can use similarity to prove the statement. Note that the question is asking the "difference of $AX$ and $BY$", this means absolute value of the difference as you can see below.
Suppose, $OP$ passes through $AC$ at $M$. We have triangles $AMX$, $CMZ$, and $BMY$ all similar to each other. By similarity $$\frac{AX}{CZ}=\frac{AC-MC}{MC}, \frac{BY}{CZ}=\frac{BC+MC}{MC}$$
Since $AC=BC$, subtracting the two we get the result.