# Proving that $ABC$ is similar to $DQP$

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?

• you can show that $PQ \parallel BC$ by looking at the pair of similar triangles $MGP, BPQ$ and $GQN, BQC.$ – abel Jun 21 '15 at 15:21
• @abel hmm I see thank you! But what about the collinearity of E, P and D? – WilliamKin Jun 21 '15 at 16:21
• where/how did you define the point $E?$ – abel Jun 21 '15 at 16:24
• @abel oh right I'm so sorry for not clarifying! E is the midpoint of AB and F is the midpoint of AC – WilliamKin Jun 21 '15 at 16:50
• @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. – Blue Jun 21 '15 at 21:04

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.