In a triangle with $\overline{AB}=62$ , $\overline{AC}=50$ and $\overline{BC}=105$, find the lenght of the segment $\overline{Q_1Q_2}$ Recently, I have found this problem:

In a triangle $\triangle{ABC}$, the side's lenghts are: $\overline{AB}=62$ , $\overline{AC}=50$ and $\overline{BC}=105$. Let $M$ the mid point of the segment $AB$ and let $N$ the mid point of the median. Now, let $P_1$ the mid point of $CN$ and let $P_2$ the mid point of $NM$. Finally, let
respectively $Q_1$ and $Q_2$ the point of intersection of $BC$ with the lines $AP_1$ and $AP_2$. Find the lenght of the segment $Q_1Q_2$.

In order to explain in the clearest way the problem, I made this graph:

Here the hypotesis:

*

*$\overline{AB}=62$ , $\overline{AC}=50$ and $\overline{BC}=105$.

*$\overline{AM}=\overline{MB}$.

*$\overline{MN}=\overline{CN}$.

*$\overline{MP_2}=\overline{P_2N}=\overline{NP_1}=\overline{P_1C}$.

Here the goal: find the lenght of $\overline{Q_1Q_2}$.
I have solved this problem using the cartesian plane. The calculations are very long and it would require a very long time to write them here, so I decided to describe only the strategy:

*

*Let $C(0,0)$ and $B(105,0)$. Find the coords of the popint $A\left(\frac{461}{19},y_A\right)$ where $y_A$ is a complicated number.

*Find the coords of $M, P_1\left(\frac{1511}{80},\frac{y_A}{8}\right) \; \text{and} \; P_2\left(\frac{4533}{80},\frac{3y_A}{8}\right)$.

*I find the intersection of the line $AP_1$ and $AP_2$ with the $x-$axis. So, I have the coords of the points $Q_1(15,0)$ and $Q_2(63,0)$.

I would like to have a geometrical solution to this problem. With trigonometry, I could obtain all the information about the sides and the angles, but the numbers, especially sines and cosines, are very complicated. Are there any other methods?
 A: There is a very simple solution through Ceva and Van Obel's theorems.
Let $R_2=AC\cap BP_2$ and $R_1=AC\cap BP_1$. By Ceva's theorem $Q_1 R_1\parallel Q_2 R_2\parallel AB$.
By Van Obel's theorem
$$ \frac{1}{3}=\frac{CP_1}{P_1 M}= 2\frac{CQ_1}{Q_1 B},\qquad 3=\frac{CP_2}{P_2 M}=2\frac{CQ_2}{Q_2B}. $$
It follows that $CQ_1:CB=1:7$ and $CQ_2:CB=3:5$, so
$$ Q_1 Q_2 = \left(\frac{3}{5}-\frac{1}{7}\right) BC=\color{red}{48}.$$
A: Multiple use of the cosine rule should work:
$a^2 = b^2 + c^2 - 2 b c \cos A$
Do the following in sequence:
Use Cosine rule to get $\angle BAC$
Use Cosine rule to get $MC$
Use Cosine rule to get $\angle MAP_2$ and $\angle P_1AC$
Use Cosine rule to get $BQ_2$
Use Cosine rule to get $\angle CQ_1$
and you have what you need.
HINT: Don't plug in the numbers till you have worked out the algebra. One of the commonest idiocies in mathematics is to start with the numbers. Do the algebra first, in order to get the knowledge of the relationships, then, and only then, do you start mucking about with arithmetic.
A: This is a sketch of a solution, numerical figures will not be plugged in.
First, by Apollonius's theorem, the length of the median $\overline {CM}$ may be calculated by $\overline {CA}^2+\overline {CB}^2=2(\overline {MC}^2+\overline {MB}^2)$. Construct a line passing through C and parallel to $\overline {AB}$. Extend $\overline {AQ_1}$ and let it intersect the line constructed before at R, then $\overline {CR}: \overline{MA}=\overline{MP_1}: \overline{P_1C}$, so the length of $\overline {CR}$ may also be found. Additionally, $\overline {CR}: \overline{BA}=\overline {CQ_1}: \overline{BQ_1}$.
The area of the whole triangle may be found via Heron's formula. Since $N, P_1, P_2$ divide  $\overline {CM}$ evenly,  the areas of $AP_2M, AP_2N$, etc. are the same may be explicitly computed. We may use again similar triangles to find the area of $CP_1R$, and the areas of $BQ_1A, CQ_1R$ by $\overline {CQ_1}:\overline{BQ_1}$. Then we have the relation among the areas $CQ_1P_1=CP_1R-CQ_1R$, which allows us to find the area of $CQ_1P_1$ and thus $\overline{P_1Q_1}:\overline{P_1A}$.
Connect $\overline{BP_1}$ and let it meet $\overline{AQ_2}$ at S. In triangle $P_1AB, \overline {P_1M}$ is a median and $\overline {P_1P_2}:\overline {P_2M}=2:1$, so $P_2$ is the center of gravity. Consequently, $\overline {P_2S}:\overline {P_2A}=1:2$ and $\overline {P_1S}=\overline {BS}$. By Menelaus' theorem, $\dfrac{\overline{BQ_2}}{\overline{Q_1Q_2}}\dfrac{\overline{AQ_1}}{\overline{AQ_2}}=1$. Having found $\overline{BQ_2}:\overline{Q_1Q_2}$, we may compute the length of $\overline {Q_1Q_2}$ from the length of $\overline{BC}$, and thus the  desired length is found.
A: Let $M_2$ be the intersection point of $BC$ with the parallel to $AQ_2$ passing through $M$.
Applying Thales’ theorem to the parallel straight lines   $M_2M$ and $Q_2A$ cutting $BC$ and $MC$, we obtain that:
$\frac{M_2Q_2}{Q_2C}=\frac{MP_2}{P_2C}$. $\;\;\;(1)$
Applying Thales’ theorem to the parallel straight lines   $M_2M$ and $Q_2A$  cutting $BC$ and $BA$, we obtain that:
$\frac{BQ_2}{M_2Q_2}=\frac{BA}{MA}$. $\;\;\;(2)$
By multiplying $(1)$ and $(2)$ side by side, we obtain that:
$\frac{BQ_2}{Q_2C}=\frac{MP_2}{P_2C}\cdot\frac{BA}{MA}$
$\frac{BQ_2+Q_2C}{Q_2C}=\frac{MP_2}{P_2C}\cdot\frac{BA}{MA}+1$
$\frac{BC}{Q_2C}=\frac{MP_2}{P_2C}\cdot\frac{BA}{MA}+1.\;\;\;(3)$
Analogously it is possible to obtain the following equality:
$\frac{BC}{Q_1C}=\frac{MP_1}{P_1C}\cdot\frac{BA}{MA}+1.\;\;\;(4)$
By $(3)$ and $(4)$ we get:
$\frac{BC}{Q_2C}=\frac{1}{3}\cdot 2 +1 =\frac{5}{3}$
$\frac{BC}{Q_1C}=3\cdot 2 +1 =7$.
Therefore:
$\frac{Q_2C}{BC}=\frac{3}{5}$
$\frac{Q_1C}{BC}=\frac{1}{7}$
and, by subtracting the previous equalities side by side, we obtain that:
$\frac{Q_2Q_1}{BC}=\frac{3}{5}-\frac{1}{7}=\frac{16}{35}$
$Q_2Q_1=\frac{16}{35}\cdot BC$.
A: This is a direct explanation of Menelaus' theorem. It says, that if you can draw a line (called a $\textit{transversal}$) through the sides of the $\Delta ABC$ intersecting its sides $AB,BC,CA$, possibly extended, at $F,D,E$ respectively, then $$\dfrac{BD}{DC}\dfrac{CE}{EA}\dfrac{AF}{FB} = -1$$ where the negative sign is due to directed segments, according to the notation $AF=-FA$ and so on. If we are dealing only with lengths, as we are doing here, then it suffices to work with the product being equal to $1$.
Now, we first apply the theorem to $\Delta CBM$ with transversal $Q_2P_2A$, which gives
$$\dfrac{BA}{AM}\dfrac{MP_2}{P_2C}\dfrac{CQ_2}{Q_2B} = 1 \implies \dfrac21\dfrac13 \dfrac{CQ_2}{Q_2B}=1 \implies CQ_2 = \dfrac32 Q_2B \\ \implies CQ_2+Q_2B = \dfrac52Q_2B \implies Q_2B=\dfrac25{BC} -- (1)$$
Similarly apply the theorem to $\Delta CBM$ with transversal $Q_1P_1A$, which gives $$\dfrac{BA}{AM}\dfrac{MP_1}{P_1C}\dfrac{CQ_1}{Q_1B} = 1 \implies \dfrac21\dfrac31 \dfrac{CQ_1}{Q_1B}=1 \implies BQ_1 = 6CQ_1 \\ \implies BQ_1+CQ_1 = 7CQ_1 \implies CQ_1=\dfrac17{BC} -- (2)$$
Combining $(1)$ and $(2)$, we get,
$$\therefore Q_1Q_2 = BC - BQ_2 - CQ_1 = BC - \dfrac25 BC - \dfrac17 BC = \dfrac{16}{35}BC  = 48$$
