0
$\begingroup$

Let $G$ be the centroid of triangle $ABC$. Let $D$ be the midpoint of $BC$. A line through $G$ parallel to $BC$ meet $AB$ at $M$ and $AC$ at $N$. $MC$ meets $BG$ at $P$ and $NB$ meets $CG$ at $Q$. Prove that triangle $DQP$ is similar to triangle $ABC$.

I observed, by drawing an accurate diagram, that $E, P, D$ seem to be collinear and so are $D, Q, F$. It is obvious that the triangle $DEF$ is similar to $ABC$, and hence I was thinking of proving that the points mentioned above are collinear, AND $PQ//EF$ hence triangle $DPQ$ is similar to $DEF$ which is then similar to $ABC$. How should I go about proving this?

$\endgroup$
  • $\begingroup$ you can show that $PQ \parallel BC$ by looking at the pair of similar triangles $MGP, BPQ$ and $GQN, BQC.$ $\endgroup$ – abel Jun 21 '15 at 15:21
  • $\begingroup$ @abel hmm I see thank you! But what about the collinearity of E, P and D? $\endgroup$ – WilliamKin Jun 21 '15 at 16:21
  • $\begingroup$ where/how did you define the point $E?$ $\endgroup$ – abel Jun 21 '15 at 16:24
  • $\begingroup$ @abel oh right I'm so sorry for not clarifying! E is the midpoint of AB and F is the midpoint of AC $\endgroup$ – WilliamKin Jun 21 '15 at 16:50
  • $\begingroup$ @WilliamKin: (I would've written $E$ is the midpt of $CA$, and $F$ the midpt of $AB$, but I'll keep your notation.) In $\triangle FBC$, cevians $CG$ and $BN$ meet at $Q$; let $FX$, with $X$ on $BC$ be the cevian from $F$ through $Q$. By Ceva's Theorem, $$\frac{|CN|}{|NF|} \cdot \frac{|EG|}{|GB|} \cdot \frac{|BX|}{|XC|} = 1$$ Note that centroid $G$ trisect the medians, so that $|FG|/|GB| = 1/2$; also, $|CN| = \frac13|AC|$ (why?), so that $|NF| = \frac16|AC|$, which gives $|CN|/|NF| = 2$. Thus, $|BX|/|XC| = 1$, which implies that $X$ coincides with $D$: points $F$, $Q$, $D$ are indeed collinear. $\endgroup$ – Blue Jun 21 '15 at 21:04
0
$\begingroup$

Refer to the following figure:

enter image description here

To prove $E,P$ and $D$'s collinearity, you may notice that $\triangle GPM$ and $\triangle BPC$ are similar. $\triangle GPM$ can be transformed into $\triangle BPC$ by rotating $180^\circ$ (and enlarging), and vice versa $\color{red}{\left(1\right)}$. Hence their medians from $P$ form a straight line. $MG\parallel BC$, and then $ED$ cuts $MG$ and $BC$ in half $\color{red}{\left(2\right)}$. The medians and $ED$ pass through the midpoints of $MG$ and $BC$, and only one straight line can pass through two points (on a plane). Therefore, the medians and $ED$ are the same straight line, and $E,P$ and $D$ are collinear.

$\endgroup$

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.