# Thread: Help with two proofs involving transversals

1. ## Help with two proofs involving transversals

Let a triangle with the sides a, b, c be given with three vertex transversals that pass through a point. Show that if for each of the three vertex transversals, one forms the quotient of their upper section ${t_{a}}^0$ and the whole segment $t_a$, then the sum of the three quotients has the constant value 2:
$\frac {{t_{a}}^0} {t_b} +\frac {{t_{b}}^0} {t_b}+\frac {{t_{c}}^0} {t_c} = 2$

I am stuck on what is meant by the "upper section." Is it referring to the segment past the point on the side of the triangle that the transversal passes through?

I appreciate the help.

2. Hello ordinalhigh
Originally Posted by ordinalhigh
Let a triangle with the sides a, b, c be given with three vertex transversals that pass through a point. Show that if for each of the three vertex transversals, one forms the quotient of their upper section ${t_{a}}^0$ and the whole segment $t_a$, then the sum of the three quotients has the constant value 2:
$\frac {{t_{a}}^0} {t_b} +\frac {{t_{b}}^0} {t_b}+\frac {{t_{c}}^0} {t_c} = 2$

I am stuck on what is meant by the "upper section." Is it referring to the segment past the point on the side of the triangle that the transversal passes through?

I appreciate the help.
I think that the question can be re-written like this:
$P$ is any point in the interior of a triangle $ABC$. Transversals $AP$, $BP$ and $CP$ are produced to meet the opposite sides at points $L$, $M$ and $N$ respectively. Prove that:
$\frac{AP}{AL}+\frac{BP}{BM}+\frac{CP}{CN}= 2$
The reason I think this is correct is that when $P$ is the centroid of the triangle, each of the ratios is equal to $\tfrac23$, and the sum is therefore $\tfrac23+\tfrac23+\tfrac23=2$.

I think that you might be able to prove it using Ceva's Theorem, but so far I'm afraid I've drawn a blank.

3. Hello ordinalhigh

I have a proof that does indeed use Ceva's Theorem. Here's the outline.

Look at the attached diagram. We have to prove that:
$\frac{AP}{AL}+\frac{BP}{BM}+\frac{CP}{CN}= 2$
Let $L,\; M,\; N$ divide $BC,\; CA,\; AB$ in the ratios $p:1,\; q:1,\; r:1$. Then, as shown in the diagram:
$AN = \frac{rc}{1+r}$

$NB = \frac{c}{1+r}$

... etc
Ceva's Theorem states that:
$\frac{AN}{NB}\cdot\frac{BL}{LC}\cdot\frac{CM}{MA}= 1$

$\Rightarrow pqr=1$
...(1)
Next, we construct the line $LQ \parallel BA$, meeting $CN$ at $Q$ and, using similar triangles, we prove that:
$LQ = \frac{BN.LC}{BC}=\frac{AN.PL}{AP}$
and hence that
$\frac{PL}{AP}= \frac{BN.LC}{AN.BC}$

$\Rightarrow 1+\frac{PL}{AP}= 1+\frac{BN.LC}{AN.BC}$

$\Rightarrow \frac{AL}{AP}=\frac{AN.BC+BN.LC}{AN.BC}$
$= ...$ etc

$= \frac{1+r+rp}{r+rp}$
$\Rightarrow \frac{AP}{AL}=\frac{r+rp}{1+r+rp}$
Similarly:
$\frac{BP}{BM}=\frac{p+pq}{1+p+q}$
and:
$\frac{CP}{CN}=\frac{q+qr}{1+q+qr}$
$\frac{AP}{AL}+\frac{BP}{BM}+\frac{CP}{CN}=\frac{r+ rp}{1+r+rp}+\frac{p+pq}{1+p+pq}+\frac{q+qr}{1+q+qr }$
$r = \frac{1}{pq}$
$\Rightarrow \frac{AP}{AL}+\frac{BP}{BM}+\frac{CP}{CN}=\frac{\d frac{1}{pq}+\dfrac{1}{q}}{1+\frac{1}{pq}+\frac{1}{ p}}+\frac{p+pq}{1+p+pq}+\frac{q+\dfrac{1}{p}}{1+q+ \dfrac{1}{p}}$
$= ...$ etc
$=2$