Showing $\frac{|\vec{A_1B}|}{|\vec{A_1C}|}\frac{|\vec{B_1C}|}{|\vec{B_1A}|}\frac{|\vec{C_1A}|}{|\vec{C_1B}|}=1$ implies $AA_1$, $BB_1$, $CC_1$ concur

Question

Hello I am trying too hard but still can't solve this

Let's assume a random triangle $\triangle ABC$ and points $A_1$, $B_1$, $C_1$ on edges $BC$, $AC$, $BA$, respectively.
Known that: $$ \frac{|\overrightarrow{A_1B}|} {|\overrightarrow{A_1C}|} \frac{|\overrightarrow{B_1C}|} {|\overrightarrow{B_1A}|} \frac{|\overrightarrow{C_1A}|} {|\overrightarrow{C_1B}|} =1 $$ Show with vector analysis that $\overleftrightarrow{AA_1}$, $\overleftrightarrow{BB_1}$, $\overleftrightarrow{CC_1}$ are concurrent lines.

First of all I draw the problem information in paper as shown in Figure 1 and I think my goal is to show that $$|{\overrightarrow{AA_1}}\, {\overrightarrow{BB_1}}\,{\overrightarrow{CC_1}}| = 0$$

I use a theorem shown in Figure 2.

If $$ \frac{\overrightarrow{AM}} {\overrightarrow{MB}} =k$$ then $$\overrightarrow{OM}=\frac{\overrightarrow{OA}+k \overrightarrow{OB}}{1+k}$$

let's say that $$ k_a=\frac{\mathop {CA_1}\limits^ \to } { \mathop {A_1B}\limits^ \to},k_b=\frac{\mathop {AB_1}\limits^ \to } { \mathop {B_1C}\limits^ \to} , k_c=\frac{\mathop {BC_1}\limits^ \to } { \mathop {C_1A}\limits^ \to}$$ then I get $$ \mathop {AA_1}\limits^ \to=\frac{\mathop {AB}\limits^ \to +k_a \mathop {AC}\limits^ \to} {1+k_a},\mathop {BB_1}\limits^ \to=\frac{\mathop {BA}\limits^ \to +k_b \mathop {BC}\limits^ \to} {1+k_b},\mathop {CC_1}\limits^ \to=\frac{\mathop {CB}\limits^ \to +k_c \mathop {CA}\limits^ \to} {1+k_c} $$

by multiplication I get $$ \mathop {AA_1}\limits^ \to \mathop {BB_1}\limits^ \to \mathop {CC_1}\limits^ \to =\frac{(\mathop {AB}\limits^ \to +k_a \mathop {AC}\limits^ \to)(\mathop {BA}\limits^ \to +k_b \mathop {BC}\limits^ \to)(\mathop {CB}\limits^ \to +k_c \mathop {CA}\limits^ \to)} {(1+k_a)(1+k_b)(1+k_c)} $$

I think almost there because denominator is positive because k are positives so numerator should be zero to achieve my goal.After numerator multiplication I get
$\mathop {AB}\limits^ \to \mathop {BA}\limits^ \to \mathop {CA}\limits^ \to k_a k_c $ $+\mathop {AB}\limits^ \to \mathop {BA}\limits^ \to \mathop {CB}\limits^ \to k_a $ $+\mathop {AB}\limits^ \to \mathop {BC}\limits^ \to \mathop {CA}\limits^ \to k_a k_b k_c $ $+\mathop {AB}\limits^ \to \mathop {BC}\limits^ \to \mathop {CB}\limits^ \to k_a k_b $ $+\mathop {AC}\limits^ \to \mathop {BA}\limits^ \to \mathop {CA}\limits^ \to k_c $ $+\mathop {AC}\limits^ \to \mathop {BA}\limits^ \to \mathop {CB}\limits^ \to $ $+\mathop {AC}\limits^ \to \mathop {BC}\limits^ \to \mathop {CA}\limits^ \to k_b k_c $ $+\mathop {AC}\limits^ \to \mathop {BC}\limits^ \to \mathop {CB}\limits^ \to k_b $

now I can't do much.
Only because $k_ak_bk_c=1$ I get

$\mathop {AB}\limits^ \to \mathop {BA}\limits^ \to \mathop {CA}\limits^ \to k_a k_c $ $+\mathop {AB}\limits^ \to \mathop {BA}\limits^ \to \mathop {CB}\limits^ \to k_a $ $+\mathop {AB}\limits^ \to \mathop {BC}\limits^ \to \mathop {CB}\limits^ \to k_a k_b $ $+\mathop {AC}\limits^ \to \mathop {BA}\limits^ \to \mathop {CA}\limits^ \to k_c $ $+\mathop {AC}\limits^ \to \mathop {BC}\limits^ \to \mathop {CA}\limits^ \to k_b k_c $ $+\mathop {AC}\limits^ \to \mathop {BC}\limits^ \to \mathop {CB}\limits^ \to k_b $

after that I can't see any continuation. Can you help me? If there is more efficient way I am happy to know. Thank you.

Answer 1

Let ABC be a triangle and suppose that P divides BC in the ratio 1 : s and that Q divides CA in the ratio 1 : t and that R divides AB in the ratio 1 : u. Then AP, BQ and CR are concurrent if and only if stu = 1.

PROOF

We can write the line through the points A and P in a parametric form with parameter alpha:

Similarly, we can write the line through the points B and Q in a parametric form with parameter beta:

Since the expressions are in the triangle coordinate form -

their coefficients will be equal at their point of intersection, thus we can equate coefficients of A and B, then solve for aplha and beta:

Solving for alpha

Similarly we work out coefficients of B and C to get:

.....10

Now we can avoid doing whole procedure again by cyclically permuting all the quantities involved. Hence we get:

.....11

Expressions 10, 11 are equal if and only if stu=1.

Hence proven