ABC is an isosceles right angle triangles having ∠C = 90°. If D is mid point on AB, then AD2 + BD2 is equal to

Answer: Option B

ΔABC is a right angle triangle.
In which ∠C = 90° and D is a point on AB such that D is perpendicular on AB.
Let AC = BC = a
∴ AB² = AC² + BC² = a² + a²
$$\begin{array}{rl}& \boxed{AB=a\sqrt{2}}\\ & \therefore BD=AD=\frac{a\sqrt{2}}{2}=\frac{a}{\sqrt{2}}\\ & \text{Now in }\Delta ACD\\ & =A{C}^2=C{D}^2+A{D}^2\\ & a^2=C{D}^2+\frac{a^2}{2}\\ & a^2-\frac{a^2}{2}=C{D}^2\\ & \boxed{\frac{a^2}{2}=C{D}^2}\\ & C{D}^2=\frac{a^2}{2}\\ & 2C{D}^2=a^2\\ & \text{and }A{D}^2+B{D}^2={\left(\frac{a}{\sqrt{2}}\right)}^2+{\left(\frac{a}{\sqrt{2}}\right)}^2\\ & =\frac{a}{\sqrt{2}}+\frac{a}{\sqrt{2}}\\ & =a^2\\ & \therefore \boxed{2C{D}^2=A{D}^2+B{D}^2}\end{array}$$