euclidean-geometrygeometry
ABCD is a rectangle with sides 10 and 20. E,F,G and H are midpoints of their sides. Find the perimeter and area of IJKL.
So I was able to figure out the sides using similar triangles ($IJ = KL = 4\sqrt2$ and $IK=JL=2\sqrt{17}$, and thus the perimeter. I am unable to find the height of the parallelogram $IJKL$.
Here is the picture for the situation:
One possibility to "see the height" is as follows.
Of course, we have an isosceles trapezium $ABCD$. In the triangle $\Delta ABC$ the sides are known, $9, 10, 17$, the semi-perimeter is $(9+10+17)/2=18$, so Heron gives its area $[ABC]=A_1+A_3$ as $$ [ABC]=A_1+A_3=\sqrt{18(18-9)(18-10)(18-17)}=36\ . $$ Since $AB=9$, the heights $DD'$ and $CC'$ of the isosceles trapezium have both length $8$. This gives then $AC'=BD'=15$, since there is the pythagorean triple $(8, 15, 17)$. We get $BC'=AD'=6$ finally. (And one can take this as a start).
Let us now denote by $A_1$ and $A_2$ the areas of the similar triangles $\Delta XAB$ and $\Delta XCD$. It is easy to compute all areas now. We may also want to compute the position of $X$, the intersection of the diagonals, on the diagonals: $$ \begin{aligned} A_1 + A_3 &= [ABC] = 36 \ ,\\ \frac {A_1}{A_3} &= \frac{XA}{XC} =\frac{AB}{CD}=\frac {9}{21}=\frac37\ ,\\ \frac {A_1}{A_1+A_3} &= \frac3{3+7}=\frac3{10}\ ,\\ A_1 &= (A_1+A_3)\cdot \frac3{10}=36\cdot \frac3{10}=\frac{54}5\ . \end{aligned} $$
$\square$
@Note: Checks, and other possibilities to get $A_1$:
The area $[ABCD]$ is $\frac 12\cdot 8(9+21)=120$, and $A_1=54/5$, $A_3=126/5$, $A_2=294/5$. We have
I will try to spend as few words as possible. The figure should be self-explanatory.
First of all notice as basic fact $$ [ADE]+[CBG]=[BAF]+[DCH]=\frac12[ABCD], $$ which implies $$ [AFCH]=[BGDE]=\frac12[ABCD]. $$
Let $M$ be the intersection of the bimedians $FH$ and $EG$. As well known and can be trivially proven $M$ bisects both bimedians.
Consider the quadrilateral $BGDE$ (first figure). Since $M$ is the midpoint of $EG$ we have: $$ \begin{cases} [BMG]=[BME]\\ [DME]=[DMG] \end{cases}\implies [BMG]+[DME]=\frac12[BGDE]=\frac14[ABCD], $$ which implies: $$ [IMJ]+[LMK]=\frac k4[ABCD]. $$ Applying the same argument to the quadrilateral $AFCH$ and summing the results one obtains: $$ [IJKL]=\frac k2[ABCD].\tag1 $$
The second figure tells us: $$ [ALE]+[BIF]+[CJG]+[DKH]=[IJKL]\tag2 $$
From the third figure we have: $$ [ABF]=2[ALE]+[BIF]+k[ABF]\implies (1-k)[ABF]=2[ALE]+[BIF]\tag3 $$
Summing the equation $(3)$ for all four triangles $ABF,BCG,CDH,DAE,$ one finally obtains: $$ (1-k)[ABCD]=3[IJKL]=\frac{3k}2[ABCD]\implies k=\frac25.\tag4 $$
Best Answer
Hint: Show that \begin{align}\operatorname{area} AJH+\operatorname{area}BFKI&=\operatorname{area} DKF+\operatorname{area}CHJL\\&=\operatorname{area}AEIJ+\operatorname{area}{CLG}\\&=\operatorname{area} DGLK+\operatorname{area}BIE=\operatorname{area}IKLJ.\end{align} For the perimeter, show that $$2\overline{LG}=2\overline{EI}=\overline{AJ}=\overline{JL}=\overline{IK}=\overline{KD}.$$ Similarly, $$2\overline{JH}=2\overline{FK}=\overline{BI}=\overline{IJ}=\overline{CL}=\overline{LK}.$$