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 3 of 3: Multiply the slopes of OQ and BC to verify perpendicularity

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In plain words

Notice that $n$ cancels out of the slope $k_{OQ}$ : the perpendicularity $QO \perp BC$ holds for any point $N$ on the angle bisector, not just the intersection with $BC$ .

k_{OQ} = \frac{\frac{m(b-c)n}{b+c} - 0}{n - n(m^2+1)} = -\frac{b-c}{(b+c)m} \implies k_{OQ} \cdot k_{BC} = -\frac{b-c}{(b+c)m} \cdot \frac{m(b+c)}{b-c} = -1

Detailed analysis

Using $Q = \left(n, \frac{m(b-c)n}{b+c}\right)$ and $O = (n(m^2 + 1), 0)$ , the slope of line $OQ$ is $k_{OQ} = \frac{\frac{m(b-c)n}{b+c}}{-n m^2} = -\frac{b-c}{(b+c)m}$ , where $n$ cancels out completely. Multiplying this by the slope $k_{BC} = \frac{m(b+c)}{b-c}$ of side $BC$ gives $k_{OQ} \cdot k_{BC} = -1$ , proving that $QO \perp BC$ .