Problem 3

Let

ABC

be a triangle. Let

M

and

N

be the points in which the median and the angle bisector, respectively, at

A

meet the side

BC

. Let

Q

and

P

be the points in which the perpendicular at

N

NA

meets

MA

and

BA

, respectively, and

O

the point in which the perpendicular at

P

BA

meets

AN

produced. Prove that

QO

is perpendicular to

BC

Step 1 of 3: Set up coordinates with AN on the x-axis and find the slopes of BC and AM

Previous step Next step

In plain words

Aligning the coordinate axes with the angle bisector makes the equations of $AB$ and $AC$ sign-reversals of each other ( $y = \pm mx$ ), stripping away all trigonometric clutter.

A = (0,0), \quad B = (b, mb), \quad C = (c, -mc), \qquad k_{BC} = \frac{m(b+c)}{b-c}, \quad k_{AM} = \frac{m(b-c)}{b+c}

Detailed analysis

Choose Cartesian coordinates with $A = (0, 0)$ and ray $AN$ along the positive $x$ -axis. Then lines $AB$ and $AC$ are symmetric across the $x$ -axis with equations $y = mx$ and $y = -mx$ , so we can write $B = (b, mb)$ and $C = (c, -mc)$ . If $b = c$ , then $AB = AC$ and $Q, N, M$ coincide, making the claim immediate; assume $b \neq c$ . Then $BC$ has slope $k_{BC} = \frac{mb - (-mc)}{b - c} = \frac{m(b+c)}{b-c}$ , and since $M = \left(\frac{b+c}{2}, \frac{m(b-c)}{2}\right)$ , the median $AM$ has slope $k_{AM} = \frac{m(b-c)}{b+c}$ .