The problem:
So apparently this is the problem.
I've "solved" this problem, but my answer doesn't match up with the book's answer. Here's how I solved it:
First of all, Triangles AOP, IOD, and ZOT are similar based on A.A
We have TZ/AP = 28/64 And AZ=46; so 46=AI+IO+OZ And also this: IO/AO can be rewritten as (IO)/(AI+IO)
Also, triangles AZP and IZE are similar. So IZ/AZ can be rewritten as (IO+OZ)/46; Also, ZE/ZP = 1/2
So (IO + OZ)/46 = 1/2 Thus we get the equation IO + OZ = 23 Plugging this value in the AZ equation, we get 46 = 23 + AI Therefore AI is also equal to 23.
Plugging AI's value in the IZ/AZ equation, we get: IO/(23+IO) = 64/28
But the solution that I am getting is 161/9 while the book's is 9. I can't seem to find what I am doing wrong.
This is how the book has done it: (Ignore the figure on the right)
Can someone point out what on earth did I do wrong? I've literally re-read my solution like 5 times and I can't seem to find a mistake. Even many of the details in my solution and the book's solution match up -- crucial details such as AO = 32, it can only be 32 when AI = 23 because based on the book, IO = 9, so 23+9 =32. Even the IZ part in my solution and the book's is the same. So it appears to be a problem with the ratio IO/(23+IO)=64/28. I cant seem to figure out what the problem is.
(The problem comes will beautiful letters from all places of the alphabet. This is maybe the main reason for my typos below. There is no scheme to remember them.)
First of all, $RE$ is the mid line in the trapez, so $I,D$ are the mid points of the diagonals, there is no reason to not go straightforward to $AI=IZ=\frac 12AZ=23$.
Now $RD, RI$ are mid lines in $\Delta TAP$, and $\Delta ATZ$ respectively, so they have lengths $\frac 12 64=32$ and $\frac 12 28=14$. This gives $ID=32-14=18$.
Now let us go to the error. You claim at some point: $$ \frac{OI}{OA} = \frac{OI}{23+OI} \overset?= \frac{64}{28} =\frac{AP}{TZ} \ . $$ Well, the starting point is $<1$ and the end is $>1$. Something is wrong already, we would get a negative $OI$. But ok, let us take the value leading to $161/9$, so: $$ \frac{OI}{OA} = \frac{OI}{23+OI} \overset?= \frac{28}{64} =\frac{TZ}{AP}\ . $$ The question mark marks the error. Note that from the similarity $$ \boxed{ \Delta OID\sim\Delta OAP } $$ we get instead: $$ \frac{OI}{23+OI} = \boxed{ \frac{OI}{OA} = \frac{\color{blue}{ID}}{AP} } = \frac{\color{blue}{18}}{64}\ . $$ Now we solve $64\cdot OI=18\cdot 23+18\cdot OI$ getting $OI=18\cdot23/(64-18)=18\cdot23/46=18/2=9$.
Note: It is good to remember. The mid line in the trapez, in our case $RE$, is half the sum of the parallel sides (lenths). The segment delimited by the diagonals from the mid line is half the difference of them.