9514 1404 393
Explanation:
X is the midpoint of AC, so will have coordinates ...
X = (A +C)/2 = ((0, 2a) +(2a, 0))/2 = (2a, 2a)/2
X = (a, a)
Then the length of AX is ...
d = √((x2 -x1)² +(y2 -y1)²)
d = √((0 -a)² +(2a -a)²) = √(2a²) = a√2
and the length of BX is ...
d = √((0 -a)² +(0 -a)²) = √(2a²) = a√2
The lengths AX and BX are the same, so ΔAXB is isosceles.
