Hexagon Area

As, $D,E,F$ are midpoints of sides of triangle $ABC$ , then the four triangles $AFE,FBD,EDC,EDC$ are congruent with area of $(1/2)^2 = 1/4$ times of triangle $ABC$ and the first three triangles can be obtained by translation of the other.

This tells us that $[EFP]+[FDQ]+[DER]$ is equal to area of any ot those triangles, which is $1/4 * 468 = 117$ . And clearly $[DEF] = 117$ too.

Thus, the area of the hexagon $[DREPFQ] = [EFP]+[FDQ]+[DER]+[DEF] = 234$

Since D,E,F are midpoints of each triangle's sides, it can be shown that the triangle AFE, EDC, FBD and DEF are all congruent. Therefore we have that [DEF] = [ABC]/4 = 468/4 = 117. Now consider triangle AFE, FBD and EDC. Since they're all congruent, small triangles obtained by joining incenter with each vertex of triangle are also congruent (i.e. [FPA] = [BQF] = [DRE]). Hence [DREPFQ] = [DEF]+[FQD]+[DRE]+[EPF] = [DEF]+[FQD]+[BQF]+[DQB] = [DEF]+[BDF] = 2[DEF] = 2*117 = 234

Suppose $I$ is the incentre of $DEF$ . Note that the four smaller triangles inside $ABC$ are all congruent to one another. So now in triangles $FQD$ and $DIF$ , we have $\angle QFD= \frac {\angle BFD}{2}=\frac {\angle EDF}{2} = \angle IDF$ , and similarly $\angle QDF = \angle IFD$ . Also $FD=DF$ , proving that $FQD$ is congruent to $DIF$ . So $[FQD]=[DIF]$ , and similarly $[PFE]=[IEF]$ and $[RED]=[IDE]$ . So $[DREPFQ]=[FQD]+[PFE]+[RED]+[DEF]$ = $[DIF]+[IEF]+[IDE]+[DEF]=2[DEF] = \frac {[ABC]}{2}=234$ , because since all the four smaller triangles are congruent, $[DEF]$ must be equal to $\frac {[ABC]}{4}$ .

Let $I$ be the incenter of $ABC$ . Then $AP,BQ,CR$ concurrent at $I$ . Furthermore, if we consider the homothety $\mathcal{Z}$ with the homothetic center $A$ and ratio $\frac{1}{2}$ , then we have $\mathcal{Z}(B)=F, \mathcal{Z}(C)=E$ . Hence, $\mathcal{Z}(BC)=FE$ and $\mathcal{Z}(I)=P$ . Therefore, $\mathcal{Z}(IBC)=PFE$ , then $[PFE]=\frac{1}{4}[IBC]$ . Similarly, we have $\begin{aligned} [DREPFQ] &= [DRE]+[EPF]+[FQD]+[DEF] \\ &= \frac{1}{4}\big( [IAB]+[IBC]+[ICA]+[ABC] \big) \\ &= \frac{1}{2} [ABC] = 234 \end{aligned}$

First, let's divide the triangle $ABC$ into 4 smaller triangles with the lines $DE$ , $DF$ , and $EF$ . Note that this also divides the hexagon $DREPFQ$ into 4 smaller triangles. Using Similarity of Triangles , we can derive that:

\because \left\{ \begin{array}{1 1} \frac {AE} {AF}=\frac {AC} {AB}\\ \angle CAB=\angle EAF \end{array} \right.

$\therefore \triangle AEF \sim \triangle ACB$

Because these 2 triangles are similar, we can say that $\frac {EF}{CB}=\frac {AE}{AC}=\frac {1}{2}$ and hence $EF=CD=DB$ . Following this, we can use the SAS (Side-Angle-Side) rule of Congruence of Triangles multiple times to see that $\triangle AEF \cong \triangle ECD \cong \triangle FDB \cong \triangle DFE$ and that each of this triangles has a quarter of the area of triangle $ABC$ .

Now, if we look at triangles $AEF$ , $ECD$ , and $FDB$ , we realize that since they are congruent, their incenter will be the same distance from the 3 vertices. If we overlap these triangles so that their sides line up, we can see that the 3 smaller triangles that are part of the hexagon $DREPFQ$ , $PEF$ , $QFD$ , and $RDE$ have one of their vertices as the incenter, and the other two vertices as 2 of the vertices of the congruent triangles $AEF$ , $ECD$ , and $FDB$ , each of which shares a base with the congruent triangles (i.e. $EF$ , $ED$ , and $FD$ ), none of which correspond to each other through congruence. In other words, their combined area is the equivalent of that of 1 of the 4 small triangles ( $AEF$ , $ECD$ , $FDB$ , and $DFE$ ) without any overlaps, which hence means that they also have a quarter of the area of triangle $ABC$ .

Adding the areas of triangle $DFE$ and triangles (PEF), $QFD$ , and $RDE$ , we get $\frac {1}{4}[ABC]+\frac {1}{4}[ABC]=\frac {1}{2}[ABC]=234$ , which is therefore, the area of the hexagon $DREPFQ$ .

Since $AB=2AF$ and $AC=2AE$ , then triangles $AFE$ and $ABC$ are similar. By a similar reasoning, we know that triangle $ABC$ is similar to triangles $FBD$ and $EDC$ as well. The incenter of any triangle is the intersection of its three angle bisectors. Suppose the three angle bisectors of triangle $ABC$ meet at $S$ . $P$ is the incenter of $AFE$ , so it should also lie on the angle bisector of $\angle A$ , i.e. $P$ is on $AS$ . Also, since $FP$ bisects $\angle AFE$ and $BS$ bisects $\angle ABC$ and $\angle AFE = \angle ABC$ (by similar triangles), then $\angle AFP = \angle ABS$ . Thus, $AFP$ and $ABS$ are similar triangles. In particular, since $\frac{AF}{AB}=\frac{1}{2}$ , then $\frac{[AFP]}{[ABS]} = (\frac{1}{2})^2 = \frac{1}{4}$ . By a similar reasoning, we know that $\frac{[AEP]}{[ACS]} = \frac{[CER]}{[CAS]} = \frac{[CDR]}{[CBS]} = \frac{[BDQ]}{[BCS]} = \frac{[BFQ]}{[BAS]} = \frac{1}{4}$ . So,

$\begin{aligned} [AFP]+[CER]+[BDQ] &= \frac{1}{4}([ABS]+[CAS]+[BCS]) \\ &= \frac{1}{4}[ABC] \end{aligned}$

$\begin{aligned} [AEP]+[CDR]+[BFQ] &=\frac{1}{4}([ACS]+[CBS]+[BAS]) \\ &=\frac{1}{4}[ABC]\end{aligned}$

Let $S_1 = [AFP]+[CER]+[BDQ]$ and $S_2 = [AEP]+[CDR]+[BFQ]$ .

Then, we have:

$\begin{aligned} [DREPFQ] &= [ABC] - S_1 - S_2 \\ &= [ABC] - \frac{1}{4}[ABC] - \frac{1}{4}[ABC] \\ &= \frac{1}{2}[ABC] \\ &= 234 \end{aligned}$

We first show that triangles $AFE, FBD,$ and $EDC$ are similar to triangle $ABC$ with a scale of $\frac {1}{2}$ , and hence congruent. Since $D, E$ and $F$ are midpoints of $BC, CA,$ and $AB$ , we have $DE \parallel BA$ , $EF \parallel CB$ and $FD \parallel AC$ . From Parallel Lines Property D , $AFE$ is similar to $ABC$ and scales by a factor of $\frac{1}{2}$ . Likewise, $FBD$ and $EDC$ are similar to $ABC$ by a factor of $\frac{1}{2}$ . This establishes the claim.

Now, we superimpose triangles $AEF, ECD$ and $FDB$ on top of each other. Then $P, Q$ and $R$ are going to be superimposed on each other. The incenter is within the triangle, since it is already contained within the inscribed circle. This shows that $[PEF]+[FQD]+[RDE]=[AEF]$ .

Let's consider triangle $DEF$ . Since $DE=\frac {1}{2} AB$ , $EF=\frac {1}{2}BC$ , $FD=\frac {1}{2}CA$ , it follows from Side-Side-Side that triangles $DEF$ and $ABC$ are similar by a factor of $\frac {1}{2}$ . Hence, $[DEF] = \left(\frac {1}{2}\right)^2 [ABC]$ .

Thus, $\begin{aligned} [DREPFQ] &= [DEF] + [DRE] + [EFP] + [FDQ] \\ &= [DEF] + [AFE] \\ &= \frac {1}{4} [ABC] + \frac {1}{4}[ABC] \\ &= 234. \\ \end{aligned}$

as simple logic In triangle ABC,first there are 4 triangle(AFE,DFE,BFD,DEC) which mean the area of DFE is 1/4 from triangle ABC

The question is What is [DREPFQ]? it include area of triangle DFE and area of triangle PFE,QFD,and DRE and if you just draw it on paper,you will get an idea that area PFE or QFD or DRE is 1/3 from triangle DEF or AEF or BDF or EDC...

so the conclusion is : [ABC]+[PFE]+[QFD]+[DRE]

1/4 area of ABC triangle+1/3 3 *1/4 area of ABC triangle which is 1/2 area of ABC triangle 468/2=234

The area of triangle DEF = \frac {1}{4} \times 468 and the total are of triangle PEF, RDE, QDE is \frac {1}{3} \times \frac {3}{4} \times 468 so the total area of DREPFQ is 234

Since the nature of the triangle (i.e. equilateral, scalene, isosceles) is not specified we can assume it to be an equilateral triangle (triangle ABC).

now, since $\bigtriangleup$ ABC is equilateral so points D,Eand Fdivide the figure into 4 parts of equal area.

Following similar notion point P divide $\bigtriangleup$ AFE into 3 equal parts and so for $\bigtriangleup$ BFD and $\bigtriangleup$ DEC.

so, area of hexagon DREPFQ= $\frac {1}{3}$ [AFE]+ $\frac {1}{3}$ [BFD]+ $\frac {1}{3}$ [DEC]+[FED].

and since,[AFE]=[BFD]=[DFE]=[DEC]= $\frac {1}{4}$ [ABC]

therefore [DREFPQ]= $\frac {1}{3} \times \frac {1}{4}$ [ABC]+ $\frac {1}{3} \times \frac {1}{4}$ [ABC]+ $\frac {1}{3} \times \frac {1}{4}$ [ABC]+ $\frac {1}{4}$ [ABC].

[DREFPQ]= $3 \times \frac {1}{12}$ [ABC]+ $\frac{1}{4}$ [ABC].

or [DREFPQ]= $\frac {1}{2}$ [ABC].

since [ABC]=468. so, [DREFPQ]=234.