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

Solution (By Examveda Team)

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
$$\eqalign{ & \boxed{AB = a\sqrt 2 } \cr & \therefore BD = AD = \frac{{a\sqrt 2 }}{2} = \frac{a}{{\sqrt 2 }} \cr & {\text{Now in }}\Delta ACD \cr & = A{C^2} = C{D^2} + A{D^2} \cr & {a^2} = C{D^2} + \frac{{{a^2}}}{2} \cr & {a^2} - \frac{{{a^2}}}{2} = C{D^2} \cr & \boxed{\frac{{{a^2}}}{2} = C{D^2}} \cr & C{D^2} = \frac{{{a^2}}}{2} \cr & 2C{D^2} = {a^2} \cr & {\text{and }}A{D^2} + B{D^2} = {\left( {\frac{a}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{a}{{\sqrt 2 }}} \right)^2} \cr & = \frac{a}{{\sqrt 2 }} + \frac{a}{{\sqrt 2 }} \cr & = {a^2} \cr & \therefore \boxed{2C{D^2} = A{D^2} + B{D^2}} \cr} $$