Prove: the ratio between the areas of $ABC$ and $AB'C'$ is $AB'\cdot\frac{AC'}{(AC \cdot AB)}$ 
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. 
 A: here is how I would go with a bit of dot's products. We denote by $AB$ the length of segment $AB$ and $\vec{AB}$ the vector going from $A$ to $B$
Using the $\frac{1}{2}\text{base}\times\text{height}$ formula, the area of triangle $ABC$ is given by $\left\lvert \frac{AB}{2} \left( AC - \frac{\vec{AB}\cdot \vec{AC}}{AB} \right) \right\rvert=\left\lvert \frac{AB}{2} \left(\left( \frac{\vec{AC}}{AC} - \frac{\vec{AB}}{AB} \right)\cdot \vec{AC} \right) \right\rvert$
similarly we can write the area of $AB'C'$ as $\left\lvert \frac{AC'}{2} \left(\left( \frac{\vec{AB'}}{AB'} - \frac{\vec{AC'}}{AC'} \right)\cdot \vec{AB'} \right) \right\rvert$. Now observe that the vectors $\frac{\vec{AC}}{AC} - \frac{\vec{AB}}{AB}$ and $\frac{\vec{AB'}}{AB'} - \frac{\vec{AC'}}{AC'}$ are equal, let's denote by $\vec{n}$ this vector.
Finally observe that $\frac{\vec{n}\cdot \vec{AB'}}{\vec{n}\cdot\vec{AC}} = \frac{AB'}{AC}$ since $\vec{AB'}$ and $\vec{AC}$ are co-linear.
The ratio you seek is now
\begin{align*}
\frac{\left\lvert \frac{AC'}{2} \left(\vec{n}\cdot \vec{AB'} \right) \right\rvert}{\left\lvert \frac{AB}{2} \left(\vec{n}\cdot \vec{AC} \right) \right\rvert}&= \frac{AC' \cdot AB'}{AB \cdot AC}
\end{align*}
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.
A: 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 $\dfrac{AB'\cdot AC'}{AB\cdot AC}$ 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
$$\frac{[DEF]}{[ABC]}=\frac{DE\cdot DF}{AB\cdot AC}\,.$$
This is simply because
$$[DEF]=\frac12\,DE\cdot DF\cdot\sin(\angle EDF)\text{ and }[ABC]=\frac12\,AB\cdot AC\cdot\sin(\angle BAC)\,,$$
and $$\sin(\angle EDF)=\sin(\angle BAC)\text{ as }\angle BAC=\angle EDF\text{ or }\angle BAC+\angle EDF=\pi\,.$$
If the dot symbol in the question actually means vector dot product, then the generalization takes a slightly different form.  In other words,
$$\frac{[DEF]}{[ABC]}=+\left(\frac{\overrightarrow{DE}\cdot \overrightarrow{DF}}{\overrightarrow{AB}\cdot \overrightarrow{AC}}\right)$$
if $\angle BAC=\angle EDF\neq \dfrac{\pi}{2}$.  On the other hand, if $\angle BAC+\angle EDF=\pi$ with $\angle BAC\neq \angle EDF$, then
$$\frac{[DEF]}{[ABC]}=-\left(\frac{\overrightarrow{DE}\cdot \overrightarrow{DF}}{\overrightarrow{AB}\cdot \overrightarrow{AC}}\right)\,.$$
A: Let the angle CAB be $\theta$ then Area of ABC is $\frac{1}{2} |AC| |AB| sin(\theta)$ and Area of AB'C' is $\frac{1}{2} |AB'| |AC'| sin(\theta)$.
Hence ratio is $\frac{|AC| |AB|}{|AB'| |AC'|} = \frac{|AC| |AB| cos(\theta)}{|AB'| |AC'| cos(\theta)}  = \frac{AC.AB}{AB'.AC'} $.
Hence the ratio is $\frac{AC.AB}{AB'.AC'} $.
Can u check your question again ?
