1
$\begingroup$

In a $\triangle ABC,$ If $D,E,F$ are the mid points of $BC,CA,AB$ respectively and $P,Q,R$ are the

only points on $AD,BE,CF$ such that $\displaystyle \frac{AP}{AG} = \alpha$ and $\displaystyle \frac{BQ}{BG} = \beta$ and $\displaystyle \frac{CR}{CG}=\gamma$

then ratio of $\triangle PQR$ to $\triangle ABC$ is , where $\alpha,\beta,\gamma \in (0,1)$ and $G$ is centroid of $\triangle ABC$

could some help me with this, thanks

$\endgroup$
1
$\begingroup$

It is easy to show that if two triangles $ABC$ and $ABD$ share the same base $AB$, with line $CD$ crossing line $AB$ at $H$, then their areas are in the ratio $CH/DH$. You can immediately apply that in your case, to obtain the well-known result (here and in the following I'll write $ABC$ meaning the area of triangle $ABC$): $$ ABG=BCG=CAG={1\over3}ABC. $$

For the same reason we have $$ PGQ={PG\over AG}AGQ=(1-\alpha)AGQ \quad\hbox{and}\quad AGQ={QG\over BG}AGB=(1-\beta)AGB, $$ so that in the end we get: $PGQ=(1-\alpha)(1-\beta){1\over3}ABC$. In the same way we also obtain: $QGR=(1-\beta)(1-\gamma){1\over3}ABC$ and $RGP=(1-\gamma)(1-\alpha){1\over3}ABC$.

By summing these three equalities we finally get: $$ PQR=PGQ+QGR+RGP=[(1-\alpha)(1-\beta)+(1-\beta)(1-\gamma)+(1-\gamma)(1-\alpha)]{1\over3}ABC. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.