triangle inscribed in a rectangle

Question

given a rectangle $ABCD$ how to construct a triangle such that $\triangle X, \triangle Y$ and $\triangle Z$ have equal areas.i dont know where to start. .i tried some algebra with the area of the triangles and used pythagoras theoram to find the sides of the triangle. i just need a hint. here is the picture

Answer 1

well this is my first answer so i cant upload a picture, so i am using the picture in the above answer. Consider, AB=CD=a... BD=AC=b... CF=c.. CE=d.'' so, AE=b-d, FD=a-c.. as the areas of 3 triangles(right angled) equal, a(b-d)=(a-c)b... b-d/b = a-c/a.... d/b=c/a... that gives you = a/b=c/d... so to construct take a right angled first take a point on one side and take another point on its adjacent side which accepts this propotionality, draw a triangle and then connect both the points to the point which is diametrically opposite to the right angle of the triangle and you are done. pls upvote if it is useful

Answer 2

Just make $$\frac{|DF|}{|FC|}=\frac{2}{\sqrt{5}+1},\ \frac{|AE|}{|EC|}=\frac{|FC|}{|CD|}=\frac{\sqrt{5}+1}{\sqrt{5}+3}$$

Answer 3

No need to use Pythagoras’ theorem. If you call $a$ and $b$ the segment parts on the left vertical side (top to bottom), $c$ and $d$ the segment parts on the bottom horizontal side (left to right), the conditions on areas are \begin{cases} a(c+d)=bc \\[6px] (a+b)d=bc \end{cases} Subtracting the equations yields $ac-bd=0$, so $d=ac/b$. Substituting in either equation we get $$ c(b^2-ab-a^2)=0 $$ Since the unknowns are positive, we obtain $$ b=\dfrac{1+\sqrt{5}}{2}a=\varphi a $$ Therefore also $$ c=\varphi d $$ You can't say more, because two equations can determine only two unknowns.