$ABCD$ is a square, $E$ is a midpoint of side $BC$, points $F$ and $G$ are on the diagonal $AC$ so that $|AF|=3\ \text{cm}$, $|GC|=4\ \text{cm}$ and $\angle{FEG}=45 ^{\circ}$. Determine the length of the segment $FG$.
How can I approach this problem, preferably without trigonometry?
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Ok after some calculus, I figured out to solve this problem. Let's draw the vertical V that goes through E, and call $\alpha$ and $\frac{\pi}{4}-\alpha$ the angles we get on the left and right of the 45° angle cut by V and call c the length of the side of the square. We can then do some trigonometry to get :
With a little trigo, we have 2 equations in $c$ and $tan(\alpha)$ that we can solve, which allow us to solve the whole problem. This makes me think that there is no "simple" answer without trigo :(
Edit : Well, seems there was a simple answer afterall. Really nice proof, I'm glad I was wrong !
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Let $H$ be the midpoint of $AC$ and $\angle EIC= 90^{\circ}$. We can observe that $$FH+3=HG+4,\quad FH+HG=x.$$ So we obtain $HG=\frac{x-1}2$. Since two corresponding angles are congruent; $\angle FEG =\angle EHG=45^{\circ}$ and $\angle EGF=\angle HGE$, we have that $\triangle FEG$ and $\triangle EHG$ are similar to each other. This gives $$FG:EG=EG:HG\implies EG^2 = FG\cdot HG=\frac{x(x-1)}2.$$ Now, note that $EI=\frac14 AC=\frac{x+7}4$ and $IG=IH-GH=\frac{x+7}{4}-\frac{x-1}2=\frac{9-x}{4}$. Since $\triangle EIG$ is a right triangle, by Pythagorean theorem, we find that $$ EG^2=\frac{x(x-1)}{2}=EI^2+IG^2=\frac{(x+7)^2}{16}+\frac{(9-x)^2}{16}, $$ which implies $x=5 $ or $x=-\frac{13}3$. Since $x>0$, we get $x=5$.
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Let $O$ be a midpoint of a segment $AC$.
Rotate $F$ for $90^{\circ}$ around $E$ to a new point $H$. This rotation takes $O$ to $C$.
Then $EF=EH$ and $\angle FEH = 90^{\circ}$ so $EG$ is angle bisector for $\angle FEH$ so $GH = GF = x$.
It is easy to see that $$CH = FO = {x+1\over 2}$$
Now $FHCE$ is cyclic quadrilateral since $\angle CFE = \angle CHE$ so $$\angle FCH = \angle FEH =90^{\circ}$$
Finaly we use Pythagoras theorem in triangle $\triangle CGH$:
$$ x^2 = 4^2+\Big({x+1\over 2}\Big)^2 \implies x=5$$
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Calling
$$ AF = a\\ FG=x\\ GC = b\\ EC = c\\ E = (X_0, Y_0)\\ O = \left(a+\frac x2, \frac x2\right) $$
Considering in the plane $X\times Y$ the square diagonal as the $X$ axis with $A$ at the origin, we have
$$ \frac{\sqrt 2}{2}(a+x+b)= 2c\\ x = \sqrt2 r\\ \left(X_0-\left(a+\frac x2\right)\right)^2+\left(Y_0-\frac x2\right)^2= r^2\\ X_0 = a+x+b -\frac{\sqrt 2}{2}c\\ Y_0 = \frac{\sqrt 2}{2}c $$
Here $r$ represents the radius for the circle which intersects the $X$ axis at $F, G$ such that $\angle FEG = \frac{\pi}{4}$ is the angle subtended by arc $AB$
Solving for $x,r,X_0,Y_0$ we get at
$$ \left\{ \begin{array}{rcl} x& =& \frac{1}{3} \left(b-a+2 \sqrt{a^2-2 b a+4 b^2}\right) \\ c& =& \frac{a+2 b+\sqrt{a^2-2 b a+4 b^2}}{3 \sqrt{2}} \\ r& =& \frac{b-a+2 \sqrt{a^2-2 b a+4 b^2}}{3 \sqrt{2}} \\ X_0&=&\frac{1}{2} \left(a+2 b+\sqrt{a^2-2 b a+4 b^2}\right) \\ Y_0&=&\frac{1}{6} \left(a+2 b+\sqrt{a^2-2 b a+4 b^2}\right) \\ \end{array} \right. $$
giving after substitution $x = 5$
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Let $DE$ intersect $AC$ at $G'$. Then $G'C:G'A=CE:AD=1:2$, hence $3G'C=AC$.
Let the perpendicular bisector of $BE$ intersect $AC$ at $F'$. Easy to see that $4AF'=AC$.
Note that $F'E=F'B=F'D$. Hence $F'$ is the circumcenter of $\triangle BED$, so $\angle DF'E = 2\angle DBE = 2\cdot 45^\circ=90^\circ$. It follows that $\angle F'EG'=45^\circ$.
Observe that $$\frac{AF'}{G'C} = \frac{AC/4}{AC/3}=\frac 34 = \frac{AF}{GC}.$$
If $AF'>AF$ then by the above equality we also have $CG'>CG$, hence $45^\circ = \angle FEG > \angle F'EG' = 45^\circ$, a contradiction. Similarly we rule out the possibility $AF'<AF$.
Hence $AF'=AF$, which means that $CG'=CG$, so $F'=F$ and $G'=G$. Therefore $AC = 4AF = 12 \mathrm{cm}$, and
$$FG = AC - AF - GC = 12\mathrm{cm} - 3\mathrm{cm} - 4\mathrm{cm} = 5\mathrm{cm}.$$
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Here's an attempt using the method of "have a hunch, then confirm it works", which probably follows the method by which the problem was first created. Intuiting where some missing lines ought to be can be helpful sometimes!
Begin with the following figure, with dimensions as given; we will show that filling in $z=4$ will force $y=3$, and also $x=5$, which is the answer we're looking for.
First observe that the angle at $E$ is equal to $\pi/4$ or $45^{\circ}$. [see below]
Now, because we have a bunch of similar triangles (the two meeting vertically at $G$, the two meeting vertically at $F$, and the right-angled ones along the right-hand side), we observe:
$2d/d = (x+y)/z$
$d/ (\frac{1}{3}d) = (x+z)/y$
$d/2d = (z/\sqrt{2})/(\frac{4}{3}d)$
Now if we set $z=4$, a little bit of linear algebra (or simple substitution if you will) gives us $d=3\sqrt{2}$, $y=3$, and $x=5$.
Notice how removing the continuations of the lines $EF$ and $EG$ makes everything look really hard!
Proof of the claim about the angle: This is because of a "fun fact" you may have discovered by playing around on squared graphing paper: the vectors $(2,1)$ and $(3,-1)$ (or suitable rotations of these) meet at that angle. You can prove this by reflecting one in the other using the standard formula from vector geometry: let $v=(2,1)$ and $n=(1,3)$ (which is orthogonal to $(3,-1)$). The standard reflection-in-a-line formula $v-2\frac{v\cdot n}{n\cdot n} n$ gives us
$(2,1) - 2\frac{5}{10}(1,3) = (2,1) - (1,3) = (1,-2)$
which is orthogonal to $(2,1)$; this implies that the reflection line $(3,-1)$ bisects that right angle, i.e. is at $\pi/4$ (or $45^{\circ}$) to $(2,1)$. $\square$
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Drop perpendiculars $FH, EI$ and $FJ$ and label $BH=y$, $EH=z$ and $\angle GEI=\alpha$:
$\hspace{2cm}$
Use the similarity of $\triangle FEH$ and $\triangle EIG$: $$\frac{GI}{HE}=\frac{EI}{FH} \Rightarrow \\ \frac{4-CI}{z}=\frac{CI}{y+2z}\Rightarrow \\ \frac{4-\frac{y+z}{\sqrt{2}}}{z}= \frac{\frac{y+z}{\sqrt{2}}}{y+2z}\Rightarrow \\ \frac{4\sqrt{2}-y-z}{z}=\frac{y+z}{y+2z}\Rightarrow\\ 3z^2+(4y-8\sqrt{2})z+y^2-4\sqrt{2}y=0 \stackrel{y=\frac3{\sqrt{2}}}{\Rightarrow} \\ 6z^2-4\sqrt{2}z-15=0 \Rightarrow \\ z=\frac3{\sqrt{2}}.$$ Hence: $$x+7=2(y+z)\sqrt{2}=12 \Rightarrow x=5.$$
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With the figure as labeled below, define $$a := |AA^\prime| \qquad b := |BB^\prime| \qquad c := |A^\prime B^\prime| \qquad d := |A^\prime M|$$
Note that the perpendicular from midpoint $M$ to the diagonal necessarily quadrisects that diagonal; moreover, $\overline{MM^\prime}\cong\overline{BM^\prime}$.
$$\begin{align} \triangle A^\prime M^\prime M \text{ is a right triangle} &\quad\to\quad d^2 = \left(c+b-\frac14(a+b+c)\right)^2+\left(\frac14(a+b+c)\right)^2 \tag{1}\\[4pt] \triangle A^\prime B^\prime M \sim \triangle A^\prime M B \text{ (Angle-Angle)} &\quad\to\quad \frac{|A^\prime M|}{|A^\prime B^\prime|} = \frac{|A^\prime B|}{|A^\prime M|} \quad\to\quad d^2 = c(b+c) \tag{2} \end{align}$$
Equating $d^2$ with $d^2$ yields $$3c^2 + 2c(a-b) - a^2 + 2 a b - 5 b^2 = 0 \tag{3}$$ which suggests that there's nothing inherently Pythagorean about the general values $a$, $b$, $c$ (but see the Addendum). However, for the specific values $a=3$ and $b=4$, we have $$0 = 3 c^2 - 2c - 65 = (c-5)(3c+13) \quad\to\quad c = 5 \tag{4}$$ where we have ignored the "obviously"-extraneous solution $c=-13/3$. $\square$
Note. The extraneous solution from $(4)$ becomes valid if we interpret the problem as saying that "lines $\overleftrightarrow{MA^\prime}$ and $\overleftrightarrow{MB^\prime}$ make a $45^\circ$ angle".
Addendum. I notice that there is actually something inherently Pythagorean about general $a$, $b$, $c$. We can write $(3)$ as $$\left(\frac{- a + b + c}{2} \right)^2 + b^2 = c^2 \tag{3'}$$ implying that $|-a+b+c|/2$, $|b|$, $|c|$ make a right triangle. (With $a=3$ and $b=4$, the two solutions to $(4)$ correspond to a $3$-$4$-$5$ triangle, and a $5$-$12$-$13$ triangle scaled by $1/3$.) I hadn't seen a way to make $(3')$ obvious in the figure (I was just playing with the algebra); however, by interesting coincidence, @greedoid's new answer seems to show exactly this! Here's a version of @greedoid's figure, with a more-arbitrary relationship between lengths $a$ and $b$.
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I do not like it. This is not a geometry problem; it is indeed a puzzle where you have to find this coincidence, which occurs only if $AF:GC=3:4$.
If we can find/prove that $GDF\angle=45^\circ$ in some different way, then we can define the point $P$ that as the reflection of $C$ about $DE$ and the reflection of $A$ about $DF$. Due to $FPD\angle=DPG\angle=45^\circ$, the triangle $FGP$ is right-angled and its legs are $PF=AF=3$, $PG=CG=4$, so $FG=5$.
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Without trigonometry or algebra, to find the length of $FG$.
With the figure and conditions as posted, construct circles with centers $F$, $G$ and radii $FA$, $GC$. Let the circle centered on $G$ cross $AC$, $BC$, $DC$ at $J$, $K$, $L$, and join $GK$ and $GL$.
Collinear $GK, GL$ are thus a diameter of the circle about $G$ and perpendicular to $AC$.
On $BA$ make $BM=BK$, and join $MK$, $MJ$, and $JK$.
Since $MBK$ and $ABC$ are isosceles right triangles sharing the angle at $B$ $$MK\parallel AJ$$
$FG$">
And since$$AM=CK=JK$$and$$AM\parallel JK$$then $AJKM$ is a parallelogram and$$AJ=MK=CG=4$$
Therefore$$FJ=AJ-AF=4-3=1$$ and$$FG=FJ+JG=1+4=5$$
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Draw the red lines:
$\hspace{3cm}$
The steps:
1) Draw $EO=OK$ and $KP$ parallel to $EG$ so that $\triangle CGE$ and $\triangle OPK$ are congruent, in particular, $OP=4$.
2) Draw $KM$ so that $MO=OG$, hence $\triangle KMO$ and $\triangle OGE$ are congruent (two sides and angle between them are equal, i.e. $MO=OG, KO=OE, \angle KOM=\angle EOG=45^\circ$).
3) $\triangle AKM$ and $\triangle CEG$ are congruent (because $\triangle KMO$ and $OGE$ are congruent)
4) $\triangle KPM$ is isosceles ($KP=KM=EG$).
5) $KT$ is perpendicular to $PM$ (because $\triangle AKM$ and $\triangle KTO$ are symmetric (reflective) around it.
6) If we label $PT=z$, then $AP=3-z,TM=z,MO=4-2z,OG=3-z$
7) Since $MO=OG \Rightarrow 4-2z=3-z \Rightarrow z=1$ and $x=z+4-2z+3-z=7-2z=5.$
Note: No formula (or theorem) is used.
Following solution uses no concept of concyclic or similar triangles, only rotation and Pythagoras theorem.
Let $O$ be a midpoint of a segment $AC$.
Rotate $G$ for $-90^{\circ}$ around $E$ to a new point $K$. This rotation takes $C$ to $O$ and $O$ to $B$, so $K$ is on a segment $BO$ (which is a part of diagonal $BD$)
Then $EK=EG$ so $E$ is on perpendicular bisector of $KG$. Since $\angle KEF = \angle FEG = 45^{\circ}$ line $EG$ is perpendicular bisector for $KG$ so $FK = FG = x$.
Since $ FO = {x+1\over 2}$ we can use Pythagoras theorem in triangle $\triangle FOK$:
$$ x^2 = 4^2+\Big({x+1\over 2}\Big)^2 \implies x=5$$