In a previous post, we proved an identity about triangle areas arising from reflection. In this occasion, we prove another interesting identity associated with triangle areas.
Theorem. Let $ABC$ be a triangle. Denote $D$, $E$ and $F$ arbitrary points on sides $BC$, $AC$ and $AB$, respectively. Let $A'$ be the reflection of $A$ with respect to $D$. Define $B'$ and $C'$ similarly. Denote $A''$ the reflection of $A'$ with respect to the midpoint of $BC$. Define $B''$ and $C''$ similarly. Then
$$[A'B'C']-[A''B''C''] = 3[ABC].$$
The brackets $[\, ]$ represent the area of the enclosed figure.
Lemma 1. $[A'B'C]=[ABA''B'']$, $[A'C'B]=[ACA''C'']$ and $[B'C'A]=[BCB''C'']$.
Proof. Observe that the diagonals of $BCB'C'$ and $BCEF$ are in a 2:1 ratio, and the angle between them remains unchanged. Hence, $[BCB'C']=4[BCEF]$. This is easily deduced from the formula for the area of a quadrilateral $K=\frac12pq\sin{\theta}$, where the lengths of the diagonals are $p$ and $q$, and the angle between them is $\theta$.
Also, observe that $AF$ and $BF$ are medians of triangles $\triangle{ACC'}$ and $\triangle{BCC'}$, respectively. It follows that $[ACF]=[AC'F]$ and $[BCF]=[BC'F]$, and then $[ABC]=[ABC']$. Analogously, $[ABC]=[ACB']=[BCA']$. Now, it turns out that
$$[AB'C']=[BCB'C']-3[ABC]=4[BCEF]-3[ABC]=4[ABC]-4[AEF]-3[ABC]=[ABC]-4[AEF].\tag{1}$$
Let $H$ and $I$ be the midpoints of $AC$ and $AB$, respectively. Then
$$[ABC]=4[AHI]=4[AEF]+4[HEFI].\tag{2}$$
From $(1)$ and $(2)$ follows that $[AB'C']=4[HEFI]$. Now, observe that $BC$, $BB''$, and $CC''$ are the homothetic images with a scale factor of 2 of $IH$, $EH$, and $FI$ with respect to $A$, $B'$, and $C'$, respectively. It follows that $BCC''B''$ and $HEFI$ are homothetic with $[BCC''B'']=4[HEFI]$, therefore,
$$[AB'C']=[BCC''B''].$$
$\square$
A similar reasoning must show that $[A'B'C]=[ABB''A'']$, and for the case of $\triangle{A'C'B}$ and $ACA''C''$ where $F$ and $D$ are on opposite sides of the line $GI$, where $G$ is the midpoint of the side $BC$, only a minor adjustment will be required. |
| $\color{blue}{[A'B'C]=[ABA''B'']}$, $\color{red}{[A'C'B]=[ACA''C'']}$ and $\color{green}{[B'C'A]=[BCB''C'']}$. |
Remark. Notice lemma 1 gives us an alternative way to construct a triangle with an area equal to a given quadrilateral, as long as it is not a rectangle or a square.
Back to the main problem
Notice that
$$[A'B'C']=[A'C'B'C]-[A'B'C]=4[ABC]+[AB'C']+[BA'C']-[A'B'C].$$
But $[AB'C']=[ABC]-4[AEF]$ and similarly $[BA'C']=[ABC]-4[BDF]$ and $[A'B'C]=4[CDE]-[ABC]$, so
$$[A'B'C']= 7[ABC]-4[AEF]-4[BDF]-4[CDE].\tag{3}$$
As $HIFE$ and $BCC''B''$ are inversely homothetic with scale factor $-2$, then so are segments $EF$ and $B''C''$ and similarly for $ED$ and $A''C''$ and $FD$ and $A''C''$, meaning that triangles $\triangle{DEF}$ and $\triangle{A''B''C''}$ are also inversely homothetic with factor $-2$ and then$$[A''B''C'']=4[DEF]=4[ABC]-4[AEF]-4[BDF]-4[CDE].\tag{4}$$
Substracting $(4)$ from $(3)$ we get
$$[A'B'C']-[A''B''C'']=3[ABC].$$
$\square$
