Van Khea's areal problem

 The following problem is a generalization of my previous problem conjectured by Van Khea. We will give a proof.

We will be using standard notation: $\lvert{BC}\rvert=a$, $\lvert{AC}\rvert=b$, $\lvert{AB}\rvert=c$; $\angle{BAC}=\alpha$, $\angle{ABC}=\beta$ and $\angle{BCA}=\gamma$. If $X$, $Y$ and $Z$ are the vertices of a triangle, we denote its area $[XYZ]$.

Problem. Let $ABC$ be a triangle and $P$ any point on the plane of $ABC$. Let $X$, $Y$ and $Z$ be arbitrary points on sides $BC$, $AC$ and $AB$, respectively. Let $D$ be the reflection of $P$ around $X$. Similarly, define $E$ and $F$. Denote $U$, $V$ and $W$ the midpoints of sides $BC$, $AC$ and $AB$, respectively. Let $D'$ be the reflection of $D$ around $U$. Similarly, define $E'$ and $F'$ (see figure below). Prove that

$$[DEF]+[D'E'F']=[ABC].$$

Proof. Denote $AZ=g$, $BZ=h$, $BX=j$, $CX=k$, $CY=l$ and $AY=m$. The area of ​​triangle $XYZ$ can be expressed as follows

$$[XYZ]=\frac12bc\sin{\alpha}-\frac12gm\sin{\alpha}-\frac12hj\sin{\beta}-\frac12kl\sin{\gamma}.$$

Since triangles $XYZ$ and $DEF$ are homothetic with scale factor $2$, it follows that $[DEF]=4[XYZ]$. Thus, we have

$${[DEF]}=2bc\sin{\alpha}-2gm\sin{\alpha}-2hj\sin{\beta}-2kl\sin{\gamma}.\tag{1}$$

Dividing both sides of $(1)$ by $[ABC]$ we have

$$\frac{[DEF]}{[ABC]}=4\left[1-\left(\frac{gm}{bc}+\frac{hj}{ca}+\frac{kl}{ab}\right)\right].\tag{2}$$

Segments $UX$ and $PD'$ are homothetic with center at $D$ and scale factor $2$. It follows that

$$PD'=2(j-\frac12a)=2j-a.$$

Similarly, we get $PE'=b-2l$ and $PF'=c-2g$. Since $UX$ and $PD'$ are homothetic segments, then $UX$ and $PD'$ are parallel and so are $VY$ and $PE'$. Hence $\angle{D'PE'}=\gamma$. Similarly, $\angle{D'PF'}=\beta$. So the area of triangle $D'E'F'$ is given by the expression

$$\begin{aligned}{[D'E'F']}&=[D'PE']+[D'PF']-[E'F'P]\\&=\frac{(2j-a)(b-2l)\sin{\gamma}}{2}+\frac{(2j-a)(c-2g)\sin{\beta}}{2}-\frac{(b-2l)(c-2g)\sin{(\beta+\gamma)}}{2}.\end{aligned}$$

Taking into account that $\sin{(\beta+\gamma)}=\sin{(\pi-\alpha)}=\sin{\alpha}$ and dividing by $[ABC]$ we obtain

$$\frac{[D'E'F']}{[ABC]}=\frac{(2j-a)(b-2l)}{ab}+\frac{(2j-a)(c-2g)}{ca}-\frac{(b-2l)(c-2g)}{bc}.\tag{3}$$

Adding equations $(2)$ and $(3)$, expanding and factorizing,

$$\frac{[DEF]}{[ABC]}+\frac{[D'E'F']}{[ABC]}=\frac{ca(4l+b)-4ga(m+l-b)-4j(b(g+h-c)+lc)-4klc}{abc}.$$

But $b=m+l$, $c=g+h$ and $a=j+k$, so

$$\begin{aligned}\frac{[DEF]}{[ABC]}+\frac{[D'E'F']}{[ABC]}&=\frac{ca(4l+b)-4jlc-4klc}{abc}\\&=\frac{ca(4l+b)-4cl(j+k)}{abc}\\&=\frac{ca(4l+b)-4cla}{abc}\\&=1.\end{aligned}$$

Therefore, 

$$[DEF]+[D'E'F']=[ABC].$$

$\square$

Note: The point $P$ may cross the side lines of the triangle $ABC$ in points either interior or exterior to the sides. The reasoning in cases other than that considered above requires only minor adjustments.

Remark: This theorem remains valid if $P$ is an arbitrary point in three-dimensional space.

See also

Another Identity on Triangle Areas Arising from Reflection