A CEVIAN SLOPE FORMULA WITH
APPLICATIONS

Antoine Mhanna

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Antoine Mhanna. A CEVIAN SLOPE FORMULA WITH APPLICATIONS. 2025. hal-05247409

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Distributed under a Creative Commons Attribution 4.0 International License
# A CEVIAN SLOPE FORMULA WITH APPLICATIONS

ANTOINE MHANNA

**ABSTRACT.** This note presents a formula relating the sides' slopes of a triangle with a cevian slope given a point on one side. The simple formula has many applications: from elementary geometric problems to the proof of trigonometric addition formulas.

The purpose of this note is to present a slope formula that can be used in basic coordinate geometry problems and exercises. As an application, trigonometric addition formulas are also derived.

## 1. INTRODUCTION AND A SLOPE FORMULA

We start by some basic notation:

Given an orthonormal system ($O; \vec{i}; \vec{j}$), the equation of a line ($l$) is $y = \alpha x + \beta$ for some reals $\alpha$ and $\beta$, $\alpha$ is called the slope of ($l$). In the sequel we take $\alpha$ without loss of generality to be finite and well defined; the case of a vertical line is taken as a tending limit for $\alpha \to \pm\infty$ and $\alpha = \frac{y - y_A}{x - x_A}$, for some point $A \in (l)$.

**Proposition 1.** For three non collinear points in the plan $A(x_A; y_A)$, $B(x_B; y_B)$, $C(x_C; y_C)$ and a point $M \in (AB)$ with $M(px_A + qx_B; py_A + qy_B)$ for $p, q \in \mathbb{R}^*$: $p + q = 1$, if slope (CB) = $s$, slope (CA) = $t$ and slope (BA) = $r$, then slope (CM) = $\frac{st - r.tp + sq}{sp + qt - r}$.

*Proof.* Set $f = q(y_B - y_C)$, $i = q(x_B - x_C)$, $h = p(y_A - y_C)$, $k = p(x_A - x_C)$. Consequently we have $t = \frac{h}{k}$ and $s = \frac{f}{i}$,

$$ \text{slope}(CM) = \frac{py_A + qy_B - y_C}{px_A + qx_B - x_C} = \frac{h+f}{k+i}. $$

For $x = \frac{k}{i}$, it can be verified that $r = \frac{hq - fp}{kq - ip} = \frac{txq - sp}{xq - p}$ and after simplification $x = \frac{(r - s)p}{(r - t)q}$. We also have slope (CM) = $\frac{txi + si}{xi + i}$, which is the formula given after simplifying $i$. $\square$

Notice that the numerator and the denominator in the formula can not be equal to zero simultaneously or else $s = t$, which is excluded. Say $\begin{cases} st = r.tp + sq \\ r = sp + qt \end{cases}$,

<footnote>Date: 2025.</footnote>

<footnote n="2010">2010 Mathematics Subject Classification. 51N20; 33B10; 51M04.</footnote>

<footnote>Key words and phrases. Slope of a cevian; Trigonometric addition formulas.</footnote>
$st \neq 0$, then $st = pq(s^2 + t^2) + st(p^2 + q^2)$ and $1 = pq(\frac{s^2+t^2}{st}) + (p^2 + q^2) \neq (p+q)^2$ unless $s = t$, a contradiction.

## 2. SOME APPLICATIONS

We give the following exercises as application. It should be noted of course that the problems are elementary but we illustrate how knowing the formula can shorten the calculations, if any.

### Ex.1:

Consider the points $O(0; 0)$, $E(-a; 1)$, $F(b; 1)$, $E \neq F$ for $a, b \in \mathbb{R}$.

A line ($l$) passes through $O$ of equation $y = mx$, $E \notin (l)$ and $F \notin (l)$; $(l) \cap (EF) = K$. The perpendiculars at $E$ and $F$ to $(OE)$ and $(OF)$, respectively, intersect at $C$. Finally from $H \in (l)$ let a perpendicular line to ($l$) intersect $(CF)$ at $A$ and $(CE)$ at $B$. If $(KC) \cap (HA) = M$, show that

$$M(x_M; y_M) = M(\frac{ax_A + bx_B}{a + b}; \frac{ay_A + by_B}{a + b}).$$

<figure><img src="image_2.png" /><figcaption>FIGURE 1</figcaption></figure>

Simple calculations give $K(\frac{1}{m}; 1)$ and $C(b-a; ab+1)$. We want to apply Proposition 1 to triangle $CBA$; equivalently if $t = -b$, $s = a$, $r = -\frac{1}{m}$, slope $(CK) = \frac{ab}{b-a-\frac{1}{m}}$, which is also equal to slope $(CM)$ as:

$$\frac{st - r.tp + sq}{sp + qt - r} = \frac{-ab + \frac{1}{m}(-b-\frac{a}{a+b} + a-\frac{b}{a+b})}{a-\frac{a}{a+b} - b-\frac{b}{a+b} + \frac{1}{m}} = \frac{-ab}{a-b+\frac{1}{m}}.$$

### Ex.2:
Given triangle ABC, let H be its orthocenter. In the coordinate plane, a point E = (ax<sub>B</sub> + bx<sub>C</sub>; ay<sub>B</sub> + by<sub>C</sub>) is on line (BC) with a + b = 1. Let the perpendicular to (HE) passing through C cut (AB) at point D. If S = (AH) ∩ (CD), then S = (ax<sub>D</sub> + bx<sub>C</sub>; ay<sub>D</sub> + by<sub>C</sub>).

<figure><img src="image_3.png" /><figcaption>FIGURE 2</figcaption></figure>

First apply Proposition 1 in triangle HBC, where M ≡ E, to get slope (HE):
equivalently if slope (AB) = m and slope (AC) = n, then r = 0, t = -1/n, s = -1/m,
p = a, and q = b: slope (HE) = -1/bm + an. As slope (AS) → ∞, the formula
denominator tends to zero, or we verify that in triangle ADC, slope (DC) = bm +
an, which is true.

Ex.3:

Given triangle ABC, let H be its orthocenter. In the coordinate plane, a point E = (ax<sub>B</sub> + bx<sub>C</sub>; ay<sub>B</sub> + by<sub>C</sub>) is on line (BC) with a + b = 1. Let the perpendicular to (AE) passing through A cut (CH) and (BH) at points M and N, respectively. The coordinates of point A can be written as (ax<sub>N</sub> + bx<sub>M</sub>; ay<sub>N</sub> + by<sub>M</sub>).

*Hint*: First apply Proposition 1 in triangle *ABC* to get slope (*AE*); then apply the proposition to triangle *MHN*, where slope (*HA*) → ∞.
<figure><img src="image_4.png" /><figcaption>FIGURE 3</figcaption></figure>

We note here that the exercises become more natural when choosing $p = a = q = b = \frac{1}{2}$, and then the required points are the midpoints of given segments.

Now we give the trigonometric addition formulas, see [1, 2]) for references. The idea is that the known proofs are a bit different from the one presented here. In general the addition formulas are proved, first for the sin or cos functions with ‘acute’ angles in a right triangle, and then for the tan function.

**Lemma 1.** Given two angles $a, b$ in $[0; \pi]\setminus\{\frac{\pi}{2}\}$ with $(a+b) \neq (\frac{\pi}{2} \mod \pi)$, we have the following:

$$ (2.1) \qquad \tan(a+b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} $$

*Proof.* By periodicity of the tangent function, the identity proven will be true for any real arguments $\neq (\frac{\pi}{2} \mod \pi)$, (we can also assume that $\tan(a) \to \infty$ for $a \to \frac{\pi}{2}$ in the given fraction).

To prove the identity take the following: triangle $ABC$ with $C$ as the origin and line ($CA$) as the $x$-axis; directed angles ($\vec{i}; \vec{CM}$) = $a$, $CM = 1$, and the perpendicular ($l$) to ($CM$) at $M$ cuts the $x$-axis at $A$. Finally let $B$ a point on line ($MA$) with ($\vec{CM}; \vec{CB}$) = $b$. The two different cases are given in the next figures.

Notice that case $\frac{\pi}{2} < a \le \pi$ and $\frac{\pi}{2} < b \le \pi$ can be reduced to the case $0 \le a < \frac{\pi}{2}$ and $0 \le b < \frac{\pi}{2}$ by taking the angles $\pi - a$ and $\pi - b$. We prove the first case; the other one has an identical proof.
<figure><img src="image_5.png" /><figcaption>FIGURE 4. $0 \le a < \frac{\pi}{2}$ and $0 \le b < \frac{\pi}{2}$, $MA = \tan(a)$ and $MB = \tan(b)$</figcaption></figure>

<figure><img src="image_6.png" /><figcaption>FIGURE 5. $\frac{\pi}{2} < a \le \pi$, $0 \le b < \frac{\pi}{2}$, $MA = -\tan(a)$, and $MB = \tan(b)$</figcaption></figure>

In Figure 4, apply Proposition 1: $t = 0$, $s = \tan(a+b)$, $r = -\frac{1}{\tan(a)}$, slope (CM) = $\tan(a)$, $p = \frac{\tan(b)}{\tan(a)+\tan(b)}$, and $q = \frac{\tan(a)}{\tan(a)+\tan(b)}$. $\square$

**Proposition 2.** Given two angles $a, b$ in $[-\pi; \pi]$, we have the following:

$$ (2.2) \qquad \sin(a+b) = \sin(a) \cos(b) + \sin(b) \cos(a) $$

and

$$ (2.3) \qquad \cos(a+b) = \cos(a) \cos(b) - \sin(a) \sin(b) $$

*Proof.* The cases of $a, b$ or $a+b$ equals $(\frac{\pi}{2} \mod \pi)$ are easy to verify. Also the second identity follows from the first one by taking instead of angle $a$, angle $a + \frac{\pi}{2}$. We will use Lemma 1, replacing the terms in (2.1) and simplifying:

$$ \frac{\sin^2(a+b)}{1-\sin^2(a+b)} = \frac{(\sin(a)\cos(b) + \sin(b)\cos(a))^2}{(\cos(a)\cos(b) - \sin(a)\sin(b))^2} $$

For $X = (\sin(a) \cos(b) + \sin(b) \cos(a))^2$, we have $1-X = (\cos(a) \cos(b) - \sin(a) \sin(b))^2$ by Pythagoras identity: $\sin^2(x) + \cos^2(x) = 1$. Consequently, from the expression we know that: $\sin^2(a+b) = (\sin(a) \cos(b) + \sin(b) \cos(a))^2$. It remains to show that $\sin(a+b)$ and $\sin(a) \cos(b) + \sin(b) \cos(a)$ have the same sign.
Consider $\sin(a+b) \ge 0$, the other case is given by taking opposite angles or multiplying by $-1$.

* If $\cos(a) \cos(b) > 0$, we want to prove that $\tan(a) \ge -\tan(b)$. The two cases are:

- $b$ and $a$ are in $(-\frac{\pi}{2}; \frac{\pi}{2})$, but $a \ge -b$ and since $\tan(x)$ is increasing on $[0; \frac{\pi}{2})$, we get the inequality.

- For $b$ and $a$ in $[-\pi; -\frac{\pi}{2}) \cup (\frac{\pi}{2}; \pi]$, if both angles are in $[-\pi; -\frac{\pi}{2})$ we have $\tan(a) \ge 0$ and $\tan(b) \ge 0$; if they are of opposite signs we have $0 \le a+b \le \frac{\pi}{2}$ and the inequality follows from (2.1).

* If $\cos(a) \cos(b) < 0$, we want to prove that $\tan(a) \le -\tan(b)$; $a$ in $(-\frac{\pi}{2}; \frac{\pi}{2})$ and $b \in (\frac{\pi}{2}; \pi] \cup [-\pi; -\frac{\pi}{2})$.

- For $a \ge 0$ and $b \ge 0$ we know that $a \le \pi - b$ and then $\tan(a) \le \tan(\pi - b) = -\tan(b)$.

- If $a \le 0$ and $b \ge 0$; $\tan(a) \le 0$ and $\tan(b) \le 0$.

- Finally for $a \le 0$ and $b \le 0$; $-\frac{3\pi}{2} \le a+b \le -\pi$ and the signs are given by (2.1).

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REFERENCES

[1] Ollerton R. L., A unified geometrical approach for trigonometric angle sum and difference identities. *Australian Senior Mathematics Journal*, (2018), 32(2): 32–35.

[2] Weisstein, Eric W. "Trigonometric Addition Formulas." From Mathworld–A wolfram Resource. https://mathworld.wolfram.com/TrigonometricAdditionFormulas.html.

<footnote n="1">¹ KFARDEBIAN-ZOUK, LIBRAIRIE DU LIBAN, LEBANON
Email address: tmhanat@yahoo.com</footnote>