Rectangle $ABCD$ contains a point $X$ such that $AX = 3$, $BX = 3$, and $CX = 3$. Find $DX$.

bader Dec 14, 2023

#1**0 **

Since AX=3, BX=3, and CX=3, we can extend lines AX, BX, and CX until they intersect at a point, which we'll call O.

Now let U be the foot of the perpendicular from O to line segment BD. Clearly, the sides of quadrilateral AOUX are half the diagonals of rectangle ABCD

AO=UO=DX=2AC=2AD3

We know that AC=AD3 from Pythagorean Theorem on right triangle ACD. Also, we know AD=3+3+3=9. Substituting those values, we get

AO=UO=DX=293

Therefore, DX=293.

BuiIderBoi Dec 14, 2023