Vtyjkyuk

In the accompanying figure, i am to prove that the ratio between areas $AB'C'$ and $ABC$ is $frac(AB' cdot AC')(AC cdot AB)$. Any assistance is greatly appreciated. Also, does the fact that the $B'$ and $C'$ is tangent to the inscribed circle matter here? Or could the result be generalized to any two triangles with two similar sides.

I take it that a prime symbol is missing in the question (otherwise, the question is clearly false). That is, we are supposed to show that $dfracAB'cdot AC'ABcdot AC$ is the area ratio. Use the general result below with $D:=A$, $E:=B'$, and $F:=C'$.

It is true in general that if $ABC$ and $DEF$ are triangles such that $angle BAC=angle EDF$ or $angle BAC+angle EDF=pi$, then the ratio of the area $[DEF]$ of the triangle $DEF$ by the area $[ABC]$ of the triangle $ABC$ equals
$$frac[DEF][ABC]=fracDEcdot DFABcdot AC,.$$
This is simply because
$$[DEF]=frac12,DEcdot DFcdotsin(angle EDF)text and [ABC]=frac12,ABcdot ACcdotsin(angle BAC),,$$
and $$sin(angle EDF)=sin(angle BAC)text as angle BAC=angle EDFtext or angle BAC+angle EDF=pi,.$$

If the dot symbol in the question actually means vector dot product, then the generalization takes a slightly different form. In other words,
$$frac[DEF][ABC]=+left(fracoverrightarrowDEcdot overrightarrowDFoverrightarrowABcdot overrightarrowACright)$$
if $angle BAC=angle EDFneq dfracpi2$. On the other hand, if $angle BAC+angle EDF=pi$ with $angle BACneq angle EDF$, then
$$frac[DEF][ABC]=-left(fracoverrightarrowDEcdot overrightarrowDFoverrightarrowABcdot overrightarrowACright),.$$

here is how I would go with a bit of dot's products. We denote by $AB$ the length of segment $AB$ and $vecAB$ the vector going from $A$ to $B$

Using the $frac12textbasetimestextheight$ formula, the area of triangle $ABC$ is given by $leftlvert fracAB2 left( AC - fracvecABcdot vecACAB right) rightrvert=leftlvert fracAB2 left(left( fracvecACAC - fracvecABAB right)cdot vecAC right) rightrvert$

similarly we can write the area of $AB'C'$ as $leftlvert fracAC'2 left(left( fracvecAB'AB' - fracvecAC'AC' right)cdot vecAB' right) rightrvert$. Now observe that the vectors $fracvecACAC - fracvecABAB$ and $fracvecAB'AB' - fracvecAC'AC'$ are equal, let's denote by $vecn$ this vector.

Finally observe that $fracvecncdot vecAB'vecncdotvecAC = fracAB'AC$ since $vecAB'$ and $vecAC$ are co-linear.

The ratio you seek is now
beginalign*
fracleftlvert fracAC'2 left(vecncdot vecAB' right) rightrvertleftlvert fracAB2 left(vecncdot vecAC right) rightrvert&= fracAC' cdot AB'AB cdot AC
endalign*

Which almost match your formula with a additional ' in for the nominator.

If you were talking about the dot product then please add some vec for vectors.

Let the angle CAB be $theta$ then Area of ABC is $frac12 |AC| |AB| sin(theta)$ and Area of AB'C' is $frac12 |AB'| |AC'| sin(theta)$.
Hence ratio is $fracAC' = fracAB cos(theta) = fracAC.ABAB'.AC' $.

Hence the ratio is $fracAC.ABAB'.AC' $.
Can u check your question again ?