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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.

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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 $$\frac{[DEF]}{[ABC]}=-\left(\frac{\overrightarrow{DE}\cdot \overrightarrow{DF}}{\overrightarrow{AB}\cdot \overrightarrow{AC}}\right)\,.$$

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  • $\begingroup$ Thank you! yes i accidentally missed out the apostrophe, and will edit it in the question now $\endgroup$ – Ferguson Sep 2 '18 at 10:37
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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.

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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 ?

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