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