GYanggg

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UsernameGYanggg
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 #1
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+3

The first step in order to tackle this problem is to draw another square enclosing square DEFG.

 

 

First, we prove that \(WXYZ\) is actually a square: 

\( \because\overline{ED}=\overline{DG}=\overline{GF}=\overline{FE}\because\angle{Z}=\angle{Y}=\angle{YXW}=\angle{ZWX}\because\angle{DEZ}+\angle{WEF}=90º,\angle{DEZ}+\angle{ZDE}=90º\Rightarrow\angle{ZDE}=\angle{WEF}.\)

\(\text{Using the same reasoning, we get:} \angle{ZDE}=\angle{WEF}=\angle{DGY}=\angle{GFX}.\)
\(\therefore\text{By AAS congruency:} \triangle{ZDE}\cong\triangle{YGD}\cong\triangle{XFG}\cong\triangle{WEF}.\)

 

From this, we get

\(\overline{ZE}+\overline{EW}=\overline{ZD}+\overline{DY}=\overline{YG}+\overline{GX}=\overline{FX}+\overline{FW}, \)which simplifies to \(\overline{ZW}=\overline{ZY}=\overline{YX}=\overline{XW}.\)

Therefore \(WXYZ \) is a square.

Since \(\triangle ABC\) is equilateral,\( \angle B=60º. \because \triangle BEW\) is a 30-60-90 triangle, \(\frac{\overline{EW}}{\overline{BW}}=\sqrt3.  \text{Same goes with } \triangle GXC, \frac{\overline{GX}}{\overline{XC}}=\sqrt3.\)
\(\text{If }\overline{EW}=x \text{ and } \overline{GX}=y,\text{ we get }\overline{BW}=\frac{x}{\sqrt3} \text{ and } \overline{XC}=\frac{y}{\sqrt3}.\)

If the equilateral triangle's side length is \(a, a=\overline{BW}+\overline{WF}+\overline{FX}+\overline{XC}=\frac{x}{\sqrt3}+y+x+\frac{y}{\sqrt3}.\)

After simplifying, we get \(x+y=\frac{3-\sqrt3}{2}a.\)

\(\because \overline{WX}=x+y \therefore x+y=\overline{DH}.\)

Since in any case, \(\overline{DH}=\frac{3-\sqrt3}{2}a, \) the length remains consistent.

GYanggg Nov 6, 2018
 #1
avatar+963 
+3

(a)

 

 

Let \(M\) be the midpoint of \(\overline{BC}\), connect \(A\) to \(M\). \(\overline {AM}\) must pass through \(G\) since \(G\) is the centroid and \(\overline {AM}\) is a median of \(\triangle ABC\). In addition, connect \(D\) to \(M\), \(\overline {DM}\) must pass through \(H\) since \(H\) is the centroid and \(\overline {DM}\) is the median of \(\triangle BCD\).

 

\(\because \frac{MG}{MA}=\frac13=\frac{MH}{MD} \text{ and } \angle AMD = \angle AMD \therefore\)  by SAS similarity, \(\triangle MGH \sim \triangle MAD\). Therefore\( \overline {GH} \parallel \overline {AD} \text{ and } AD=3GH.\) Using the same method to prove \(\triangle NKJ \sim \triangle NAD\), we can conclude \(\overline {KJ} \parallel \overline {AD} \text{ and } AD=3KJ. \because AD=3GH \text{ and } AD = 3KJ \therefore GH=KJ\), similarly  \(\because \overline {KJ} \parallel \overline {AD} \text{ and }\overline {GH} \parallel \overline {AD} \therefore \overline {KJ} \parallel\overline {GH}\).

 

The same method is used to prove every pair of opposite sides in the hexagon are parallel and equal in length.

 

(b)

 

 

Let the parallel line of \(\overline{HI}\) through \(G\) and the parallel line of \(\overline{GH}\) through \(I\) intersect at . \(\because \overline{GH} \parallel \overline {OI} \text { and } \overline {GO} \parallel \overline{HI} \therefore GHIO\) is a parallelogram. Then, we can conclude \([GHI]=[GOI]\). \(\because \overline{GH} \parallel \overline {OI} \text{ and } \overline {KJ} \parallel\overline {GH} \therefore \overline {OI} \parallel \overline {KJ} \because GH = OI \text { and } GH = KJ \therefore OI = KJ\).

 

From this, we can conclude \(OIJK\) is also a parallelogram, which means \([IOK]=[IJK]\). \(\because \overline{HI} \parallel \overline {GO} \text{ and } \overline {HI} \parallel\overline {LK} \therefore \overline {GO} \parallel \overline {LK} \because HI = GO \text { and } HI=LK \therefore GO = LK\). From this, we can conclude \(GOKL\) is also a parallelogram, which means \([GOK]=[GLK]\). Therefore, \([GHIJKL]=2[GIK]\). Using the same method, we can conclude \([GHIJKL]=2[LHJ]\), which means that \([LHJ]=[GIK]\).

 

I hope this helped,

 

Gavin. 

GYanggg Sep 25, 2018