Find the value of segment $x$ in the triangle below

Question

In the figure, calculate x, knowing that $a^2+b^2=36$(S:$3$)

I try

$HF=HG=HB=r$ $AC^2 = AB^2+BC^2 \implies (2R)^2=(BF+a)^2+(BG+b)^2=BF^2+2aBF+a^2+BG^2+b^2+2bBG$

$4R^2=36+BF(2a+BF)+BF^2+BG^2+BG(2b+BG)=36+BF(2a+BF+BF)+BG(BG+2b+BG)$

$4R^2 = 36+BF(2a+2BF)+BG(2b+2BG)$

$4R^2=36+2BF(a+BF)+2BG(b+BG)=36+2(AB.BF+BC.BG)$

Answer 1

Draw parallelograms as in the picture. Then it becomes clear that $x$ is the median of a right triangle with legs $a$ and $b$. Hence $x=\frac 12\sqrt{a^2+b^2}=3$.

Answer 2

Consider the $180^\circ$ rotation about $E$. Let the point $X$ be rotated to $X'$.

Steps towards a solution. If you're stuck, show what you've tried.

1. Show that $ABCB'$ is a rectangle.
   • The motivation is that this puts $a, b$ together at a right angle (EG Triangle $AFG'$), so we can use the condition $a^2 + b^2 = 36$.
2. Show that $FGF'G'$ is a parallelogram. Furthermore, $E$ is the intersection of $FF'$ and $GG'$, hence is their midpoints.

One way to show this is that $FG \parallel F'G'$, $|FG| = |F'G'|$ |GF'| = |FG'|$)

3. Hence, $ |HH'| = |FG'| = \sqrt{a^2+b^2}$.
4. Thus, $ x = |HE| = \frac{1}{2} |HH'| = \frac{ \sqrt{a^2+b^2}}{2} $.

Notes

• We could also have used a $180^\circ$ rotation about $H$, though the diagram is uglier.