Proving a formula for the area of a convex quadrilateral with an incircle

Question

I am preparing for Mathematical Olympiads. I recently stumbled across this Trigonometry Problem and I am unable to solve this after numerous attempts. I couldn't even find solution to it online.

Show that if $ABCD$ is a convex quadrilateral, that has an incircle, and $a$, $b$, $c$, and $d$ are the lengths of the tangents from the vertices to the circle, as shown then;
$$[ABCD]^2 = (a + b + c + d)(abc + abd + acd + bcd).$$

I have tried to drop altitudes from the incenter of the quadrilateral to the sides, and see if I can use some identities to prove the result. Dropping those altitudes showed that the quadrilateral was inscribed about a circle with sides tangent to the circle. I tried to use the fact that the area is equal to the semi-perimeter times the inradius. But I couldn't get any far with it because of the $r^2$ term in the R.H.S.

I am stuck at it currently. Can someone help me out with this problem? Thank You!

Answer 1

Welcome to M.S.E.

This is an amazing problem! You're going right on creating the incircle, tangents, and the altitudes. As a hint, let the four angles at the vertices be equal to $2\alpha,2\beta,2\gamma,2\delta,$ respectively. Let $O$ be the incenter. Then the lines $AO,BO, CO,$ and $DO$ would all bisect the angles of the quadrilateral, and the divided angles will have magnitudes of $\alpha, \beta, \gamma,$ and $\delta.$

Then try to apply trigonometry on the individual triangles formed by the tangency. You'll get; $$\tan{\alpha}=\frac{r}{a},\tan{\beta}=\frac{r}{b},\text{ and } \tan{\gamma}=\frac{r}{c}.$$

But then what is $\tan{(\alpha+\beta+\gamma+\delta)}$ equal to? You can use the tangent addition formula to make a cool observation; $$\dfrac{\tan(\alpha+\beta)+\tan(\gamma+\delta)}{1-\tan(\alpha+\beta)\tan(\alpha+\beta)}=\tan{(\alpha+\beta+\gamma+\delta)}.$$

It will be easy after this. I'll leave it to you to fill in the blanks. Best of Luck!

Let $I$ denote the center of the circle inscribed in quadrilateral $ABCD,$ (i.e. the incenter of quadrilateral $ABCD$). Also let $H_{AB},$ $H_{BC},$ $H_{CD},$ and $H_{DA}$ denote the foot of perpendiculars from $I$ to sides $AB,$ $BC,$ $CD,$ and $DA,$ respectively. We know that the inradius of quadrilateral $ABCD$ is$$IH_{AB}=IH_{BC}=IH_{CD}=IH_{DA}.$$Since $AH_{AB}=AH_{DA}=a,$ $IH_{AB}=IH_{BC},$ and $\angle AH_{AB}I=\angle AH_{DA}I=90^\circ,$ we know that $\triangle AH_{AB}I\cong\triangle AH_{DA}I$ by the SAS congruence theorem. Hence, we have;$$\angle H_{DA}AI=\angle H_{AB}AI.$$Similarly, we can show that $IA,$ $IB,$ $IC,$ and $ID$ bisect $\angle DAB,$ $\angle ABC,$ $\angle BCD,$ and $\angle CDA,$ respectively. We let the angles of quadrilateral $ABCD$ be $2\alpha,$ $2\beta,$ $2\gamma,$ and $2\delta;$ $$\angle DAB=2\alpha$$ $$\angle ABC=2\beta$$ $$\angle BCD=2\gamma$$ $$\angle CDA=2\delta.$$ Since $IA,$ $IB,$ $IC,$ and $ID$ bisect $\angle DAB,$ $\angle ABC,$ $\angle BCD,$ and $\angle CDA,$ respectively, we know that;$$\angle DAI=\angle IAB=\alpha$$$$\angle ABI=\angle IBC=\beta$$$$\angle BCI=\angle ICD=\gamma$$$$\angle CDI=\angle IDA=\delta.$$We know that $\angle DAB+\angle ABC+\angle BCD+\angle CDA=360^\circ,$ hence, we have;$$2\alpha+2\beta+2\gamma+2\delta=360^\circ$$$$\implies \alpha+\beta+\gamma+\delta=180^\circ.$$$$\implies \tan(\alpha+\beta+\gamma+\delta)=\tan{180^\circ}=0.$$Now we use the Tangent Sum Identity, i.e. for any two angles $x$ and $y$;$$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \tan y}.$$We have;$$\tan(\alpha+\beta+\gamma+\delta)=\frac{\tan(\alpha+\beta)+\tan(\gamma+\delta)}{1-\tan(\alpha+\beta)\tan(\gamma+\delta)}=0.$$Hence, we know that;$$\tan(\alpha+\beta)+\tan(\gamma+\delta)=0$$Applying the Tangent Sum Identity again gives us;$$\frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}+\frac{\tan \gamma + \tan \delta}{1-\tan \gamma \tan \delta}=0\tag{Equation 1}$$Analogously, we also have;$$\frac{\tan \alpha + \tan \delta}{1-\tan \alpha \tan \delta}+\frac{\tan \beta + \tan \gamma}{1-\tan \beta\tan \gamma}=0,\tag{Equation 2}$$and$$\frac{\tan \alpha + \tan \gamma}{1-\tan \alpha \tan \gamma}+\frac{\tan \beta + \tan \delta}{1-\tan \beta\tan \delta}=0.\tag{Equation 3}$$We know that;$$\tan\alpha=\frac{r}{a}$$$$\tan\beta=\frac{r}{b}$$$$\tan\gamma=\frac{r}{c}$$$$\tan\delta=\frac{r}{d}.$$Hence, we have;$$r=a\cdot\tan\alpha=b\cdot\tan\beta=c\cdot\tan\gamma=d\cdot\tan\delta.$$We know that;$$\tan\alpha+\tan\beta=\frac{r}{a}+\frac{r}{b}=\frac{r(a+b)}{ab}.$$and,$$\tan\alpha\tan\beta=\frac{r}{a}\cdot\frac{r}{b}=\frac{r^2}{ab}.$$Hence, we have;\begin{align*}\dfrac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}&=\dfrac{\frac{r(a+b)}{ab}}{1-\frac{r^2}{ab}}\\&=\dfrac{\frac{r(a+b)}{ab}}{\frac{ab-r^2}{ab}}\\&=\dfrac{r(a+b)}{ab-r^2}.\end{align*}Similarly, we also have;$$\frac{\tan \gamma + \tan \delta}{1-\tan \gamma \tan \delta}=\dfrac{r(c+d)}{cd-r^2},$$$$\frac{\tan \alpha + \tan \delta}{1-\tan \alpha \tan \delta}=\dfrac{r(a+d)}{ad-r^2},$$and$$\frac{\tan \beta + \tan \gamma}{1-\tan \beta\tan \gamma}=\dfrac{r(b+c)}{bc-r^2}.$$Now, we can substitute these values into Equation 1, and Equation 2, we have;$$\frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}+\frac{\tan \gamma + \tan \delta}{1-\tan \gamma \tan \delta}=\dfrac{r(a+b)}{ab-r^2}+\dfrac{r(c+d)}{cd-r^2}=0$$and$$\frac{\tan \alpha + \tan \delta}{1-\tan \alpha \tan \delta}+\frac{\tan \beta + \tan \gamma}{1-\tan \beta\tan \gamma}=\dfrac{r(a+d)}{ad-r^2}+\dfrac{r(b+c)}{bc-r^2}=0.$$Since in these two Equations, the two values add up to 0, we can reciprocate each value and yet keep their sum the same, we have;$$\dfrac{ab-r^2}{r(a+b)}+\dfrac{cd-r^2}{r(c+d)}=0$$and$$\dfrac{ad-r^2}{r(a+d)}+\dfrac{bc-r^2)}{r(b+c)}=0.$$Simplifying these Equations gives;$$\begin{align*} & \dfrac{(ab-r^2)(c+d)}{r(a+b)(c+d)}+\dfrac{(cd-r^2)(a+b)}{r(a+b)(c+d)}=0\\ \implies & \dfrac{(ab-r^2)(c+d)+(cd-r^2)(a+b)}{r(a+b)(c+d)}=0\\ \implies & (ab-r^2)(c+d)+(cd-r^2)(a+b)=0\\ \implies & abc+abd-(c+d)r^2+acd+bcd-(a+b)r^2=0 \end{align*}$$$$\begin{align*} \implies abc+abd+acd+bcd & =(a+b)r^2+(c+d)r^2 \\ & =(a+b+c+d)r^2 \end{align*}$$$$\implies r^2=\frac{abc+abd+acd+bcd}{a+b+c+d}.\tag{Equation 4}$$We know that;$$\begin{align*}[ABCD]&=[AIB]+[BIC]+[CID]+[DIA]\\&=\frac{1}{2}AB\cdot r+\frac{1}{2}BC\cdot r+\frac{1}{2}CD\cdot r+\frac{1}{2}DA\cdot r\\&=\frac{1}{2}(AB+BC+CD+DA)\cdot r\\&=\frac{AB+BC+CD+DA}{2}\cdot r\\&=\frac{a+b+b+c+c+d+d+a}{2}\cdot r\\&=(a+b+c+d)\cdot r.\end{align*}$$Hence, we have;$$[ABCD]^2=r^2(a+b+c+d)^2.$$By Equation 4, we know that;$$\implies r^2=\frac{abc+abd+acd+bcd}{a+b+c+d}.$$Substituting this value in Equation 4 gives us;$$\begin{align*}[ABCD]^2&=r^2(a+b+c+d)^2\\&=(a+b+c+d)^2\cdot\frac{abc+abd+acd+bcd}{a+b+c+d}\\&=(a+b+c+d)(abc+abd+acd+bcd),\end{align*}$$as desired.