Extending the Newton-Mollweide Formulas and the Law of Tangents to General Convex Quadrilaterals

 Introduction and notation

For a triangle with sides $a,b,c$ opposite angles $\alpha,\beta,\gamma$, the classical Mollweide pair is

$$\frac{a+b}{c}=\frac{\cos((\alpha-\beta)/2)}{\sin(\gamma/2)}, \qquad \frac{a-b}{c}=\frac{\sin((\alpha-\beta)/2)}{\cos(\gamma/2)}. \tag{1}$$

The triangle notation in the classical Mollweide formulas.

In 2022, the present author established a cyclic-quadrilateral generalization of Newton's form of Mollweide's formula; in 2024, the author established a cyclic-quadrilateral law of tangents. The present article extends these two earlier results to arbitrary strictly convex quadrilaterals using all four sides and both diagonals. It also derives a signed Newton–Mollweide companion identity. 

Let $ABCD$ be a strictly convex quadrilateral with distinct vertices in boundary order. Set

$$AB=a,\quad BC=b,\quad CD=c,\quad DA=d,\qquad AC=e,\quad BD=f,$$

$$\alpha=\angle DAB,\quad \beta=\angle ABC,\quad \gamma=\angle BCD,\quad \delta=\angle CDA.$$

The diagonals meet at $E$, and $\theta=\angle CED\in(0,\pi)$. All six lengths are strictly positive, all interior angles belong to $(0,\pi)$, and $\alpha+\beta+\gamma+\delta=2\pi$.

A strictly convex quadrilateral with $AC=e$, $BD=f$ and $E=AC\cap BD$; the diagonal angle is $\theta=\angle CED$.

The six-length Newton–Mollweide pair

Define the quadratic and quartic expressions

$$\begin{aligned} L_+&=e^2+f^2-a^2-c^2+2bd, &R_+&=(e^2-d^2)(f^2-b^2)+c^2((b+d)^2-a^2),\\ L_-&=a^2+c^2+2bd-e^2-f^2, &R_-&=(e^2-d^2)(b^2-f^2)+c^2(a^2-(b-d)^2). \end{aligned} \tag{2}$$

To resolve the companion's sign using lengths alone, also set

$$S=b(a^2+d^2-f^2)+d(a^2+b^2-e^2),\qquad T=d(b^2+c^2-f^2)-b(c^2+d^2-e^2). \tag{3}$$

We use $\operatorname{sgn}(0)=0$.

Theorem 1. For every strictly convex $ABCD$, $L_+>0$ and $R_+>0$, and

$$\boxed{\displaystyle \frac{\sin\frac{\alpha+\beta}{2}}{\cos((\gamma-\delta)/2)} =c\sqrt{\frac{L_+}{R_+}}.} \tag{4}$$

Moreover, $T=0$ if and only if $\gamma=\delta$. If $T\ne0$, then $R_->0$, $L_-\ge0$, and the signed six-length companion is

$$\boxed{\displaystyle \frac{\cos\frac{\alpha+\beta}{2}}{\sin\frac{\delta-\gamma}{2}} =\operatorname{sgn}(ST)\,c\sqrt{\frac{L_-}{R_-}}.} \tag{5}$$

In particular its square equals $c^2L_-/R_-$. Also, $S=0$ if and only if $\alpha+\beta=\pi$; in that case the right side of (5) is zero whenever $T\ne0$.

Proof. Write $\nu=\cos(\alpha+\beta)$. The vector identity

$$2\overrightarrow{AD}\cdot\overrightarrow{BC} =e^2+f^2-a^2-c^2$$

and the directed-turn relation $\cos(\alpha+\beta)=-\overrightarrow{AD}\cdot\overrightarrow{BC}/(bd)$ give

$$\nu=\frac{a^2+c^2-e^2-f^2}{2bd}.$$

The cosine rule yields

$$\cos\gamma=\frac{b^2+c^2-f^2}{2bc},\qquad \cos\delta=\frac{c^2+d^2-e^2}{2cd}.$$

Since $\gamma+\delta=2\pi-(\alpha+\beta)$, $\cos(\gamma-\delta)=2\cos\gamma\cos\delta-\nu$. A direct substitution into (2) now gives

$$\begin{aligned} 1-\nu&=\frac{L_+}{2bd}, &1+\cos(\gamma-\delta)&=\frac{R_+}{2bdc^2},\\ 1+\nu&=\frac{L_-}{2bd}, &1-\cos(\gamma-\delta)&=\frac{R_-}{2bdc^2}. \end{aligned}$$

Because $0<\alpha+\beta<2\pi$ and $|\gamma-\delta|<\pi$, $1-\nu>0$ and $1+\cos(\gamma-\delta)>0$. Taking the positive square root of their ratio proves (4). When $\gamma\ne\delta$, the denominator $1-\cos(\gamma-\delta)$ is positive and the second ratio yields

$$\left(\frac{\cos\frac{\alpha+\beta}{2}}{\sin\frac{\delta-\gamma}{2}}\right)^2 =c^2\frac{L_-}{R_-}. \tag{6}$$

It remains to determine the sign, which cannot be inferred from (6) alone. By the cosine rule at $A,B$, and again at $C,D$,

$$S=2abd(\cos\alpha+\cos\beta),\qquad T=2bcd(\cos\gamma-\cos\delta).$$

The identity $\cos\alpha+\cos\beta =2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$ and $\cos\frac{\alpha-\beta}{2}>0$ show that $\operatorname{sgn} S=\operatorname{sgn}\cos\frac{\alpha+\beta}{2}$. As cosine is strictly decreasing on $(0,\pi)$, $\operatorname{sgn} T=\operatorname{sgn}(\delta-\gamma)=\operatorname{sgn}\sin\frac{\delta-\gamma}{2}$. Equation (5) follows, including the zero case $S=0$. $\square$

Specialization to cyclic quadrilaterals

Put

$$\begin{gathered} s=\tfrac12(a+b+c+d),\quad X=(s-a)(s-c),\quad Y=(s-b)(s-d),\\ P=ab+cd,\quad Q=ad+bc,\quad M=ac+bd. \end{gathered} \tag{7}$$

For a strictly convex quadrilateral $s$ exceeds each side, so $X,Y,P,Q,M>0$.

Corollary 1. If $ABCD$ is cyclic, then

$$\frac{\sin\frac{\alpha+\beta}{2}}{\cos((\gamma-\delta)/2)} =\frac{a+c}{b+d}\cot\frac{\theta}{2}, \tag{8}$$

which is the cyclic Newton–Mollweide identity previously established by the present author. If $\delta\ne\gamma$ (equivalently, $b\ne d$ in the cyclic case), the signed companion is

$$\frac{\cos\frac{\alpha+\beta}{2}}{\sin\frac{\delta-\gamma}{2}} =\frac{a-c}{b-d}\tan\frac{\theta}{2}. \tag{9}$$

Proof. For a cyclic quadrilateral, the classical side–diagonal identities, which also follow from the cosine rule and the supplementary-opposite-angle relations, give

$$e^2=\frac{MQ}{P},\qquad f^2=\frac{MP}{Q},\qquad ef=M. \tag{10}$$

The definitions in (7) imply the elementary identities

$$X+Y=M,\qquad X-Y=\frac{b^2+d^2-a^2-c^2}{2}. \tag{11}$$

For any quadrilateral, the cosine rule in the four triangles with vertex $E$ gives

$$2ef\cos\theta=b^2+d^2-a^2-c^2. \tag{12}$$

Together with (10) and (11), this implies

$$\cot^2\frac{\theta}{2}=\frac{X}{Y},\qquad \tan^2\frac{\theta}{2}=\frac{Y}{X}. \tag{13}$$

Both half-angle functions are positive. Direct substitution of (10) into (2) yields

$$\begin{aligned} L_+&=\frac{4bd(a+c)^2X}{PQ},& R_+&=\frac{4bdc^2(b+d)^2Y}{PQ},\\ L_-&=\frac{4bd(a-c)^2Y}{PQ},& R_-&=\frac{4bdc^2(b-d)^2X}{PQ}. \end{aligned} \tag{14}$$

Thus (4) and (13) give (8), whereas (6) gives only the square of (9). To verify the sign, substitution into (3) further factors as

$$S=\frac{4abd(a-c)(b+d)Y}{PQ},\qquad T=\frac{4bcd(a+c)(b-d)X}{PQ}.$$

Consequently $\operatorname{sgn}(ST)=\operatorname{sgn}((a-c)(b-d))$; combining (5), (14) and (13) proves the signed identity (9). In particular $b=d\iff T=0\iff\gamma=\delta$ in the cyclic case. $\square$

A rational six-length law of tangents

The cyclic law of tangents established by the present author reads

$$\frac{\tan\frac{\alpha-\beta}{2}}{\tan\frac{\alpha+\beta}{2}} =\frac{(a-c)(b-d)}{(a+c)(b+d)}. \tag{15}$$

When $\alpha+\beta=\pi$, the tangent quotient is interpreted by its continuous extension $ (\sin\alpha-\sin\beta)/(\sin\alpha+\sin\beta)$. Define the degree-two and degree-four expressions

$$H=b^2+d^2-a^2-c^2,\quad U=e^2+a^2-b^2,\quad V=f^2+a^2-d^2, \quad P_t=2f^2U+HV,\quad Q_t=2e^2V+HU.$$

Theorem 2. For every strictly convex $ABCD$, $P_t>0$, $Q_t>0$, and

$$\boxed{\displaystyle \frac{\tan\frac{\alpha-\beta}{2}}{\tan\frac{\alpha+\beta}{2}} =\frac{bP_t-dQ_t}{bP_t+dQ_t},} \tag{16}$$

with the preceding continuous-extension convention when $\alpha+\beta=\pi$.

Proof. The sine addition/subtraction identities give, whenever the tangent quotient is defined,

$$\frac{\tan\frac{\alpha-\beta}{2}}{\tan\frac{\alpha+\beta}{2}} =\frac{\sin\alpha-\sin\beta}{\sin\alpha+\sin\beta}. \tag{17}$$

The right side is continuous for all $\alpha,\beta\in(0,\pi)$ and so defines the stated extension.

Let $x=AE$, $y=BE$, and $k=\cos\theta$. From the areas of triangles $ABD$ and $ABC$, respectively,

$$\sin\alpha=\frac{xf\sin\theta}{ad},\qquad \sin\beta=\frac{ye\sin\theta}{ab}.$$

Hence (17) equals $(bfx-dey)/(bfx+dey)$. The cosine rule in $AEB,BEC,AED$ gives

$$a^2=x^2+y^2-2xyk,\quad b^2=(e-x)^2+y^2+2(e-x)yk,\quad d^2=x^2+(f-y)^2+2x(f-y)k.$$

Subtracting the first identity from each of the others yields

$$x-ky=\frac{U}{2e},\qquad y-kx=\frac{V}{2f}. \tag{18}$$

By (12), $k=H/(2ef)$. Since $|k|<1$, solving (18) gives

$$x=\frac{2f^2U+HV}{4ef^2(1-k^2)} =\frac{P_t}{4ef^2(1-k^2)},\qquad y=\frac{2e^2V+HU}{4e^2f(1-k^2)} =\frac{Q_t}{4e^2f(1-k^2)}.$$

Because $x,y>0$, both quartics are positive. Multiplication by $bf$ and $de$, respectively, puts $bfx$ and $dey$ over the common positive denominator $4ef(1-k^2)$, proving (16). $\square$

Corollary 2. For a cyclic convex quadrilateral, the six-length law of tangents reduces to the author's previously published cyclic identity (15).

Proof. Let $K$ be the quadrilateral's area. The cyclic supplementary-angle relations imply $\sin\gamma=\sin\alpha$ and $\sin\delta=\sin\beta$. Splitting along $BD$ and along $AC$, respectively, gives

$$2K=(ad+bc)\sin\alpha=(ab+cd)\sin\beta.$$

Therefore (17) equals

$$\frac{(ab+cd)-(ad+bc)}{(ab+cd)+(ad+bc)} =\frac{(a-c)(b-d)}{(a+c)(b+d)},$$

as required. $\square$

Remark 1 (Domains and triangle degeneration). The first ratio (4) and the rational law (16) are defined on the entire strictly convex class (with the stated convention for the latter). The companion is not defined when $\delta=\gamma$; in particular, the rectangle has a $0/0$ companion and must not be included in an unqualified statement. In a cyclic degeneration $D\to C$ (so $c\to0$), use the limiting triangle $ABC$ with opposite-side lengths $\widetilde a=b$, $\widetilde b=d$, $\widetilde c=a$. The author's earlier article explains the geometric limit of the first cyclic formula: $\theta\to\angle ACB$. Under this relabelling, (8) reduces to the first formula in (1). The companion (9), when $b\ne d$, reduces to the reciprocal of the second formula in (1); equivalently, the second classical formula follows after cross-multiplication. The tangent formula (15) becomes the classical triangle law of tangents. These are limiting statements, not claims about a degenerate quadrilateral.