Prove that in any triangle the sum of the squares of any two sides is equal to twice the square - Maths - Triangles

Question

Prove that in any triangle the sum of the squares of any two sides
is equal to twice the square of half of the third side together
with twice the square of the median, which bisects the third side

Santanu Sahoo · Accepted Answer

Hi, This question needs the use of Appolonius theorem, so lets prove that to solve this, ;- Statement as per Appolonius theorem : In any triangle the sum of the squares of any two sides is equal to twice the square of half the third side together with twice the square of the median Considering the below triangle, we have:

given: ABC is a triangle and AD is the median of the triangle To Prove that : $A B^{2} + A C^{2} = 2 (B D^{2} + A D^{2})$ By construction: draw $A E ⊥ B C$ Proof: In the right angled triangle ABE : $A B^{2} = A E^{2} + B E^{2}$ (1) In the right angled triangle ACE: $A C^{2} = A E^{2} + E C^{2}$ (2) In the right angled triangle AED: $A D^{2} = A E^{2} + D E^{2} \Rightarrow A E^{2} = A D^{2} - D E^{2}$ (3) Adding eq(1) and eq(2): $A B^{2} + A C^{2} = 2 A E^{2} + B E^{2} + C E^{2} = 2 A E^{2} + (B D + D E)^{2} + (D C - D E)^{2} = 2 A E^{2} + (B D + D E)^{2} + (B D - D E)^{2} [D i s t h e m i d p o int o f B C]$ (Because DC = BD,so we substituted, BD in place of DC above) $A B^{2} + A C^{2} = 2 (A D^{2} - D E^{2}) + B D^{2} + D E^{2} + 2 B D * D E + B D^{2} + D E^{2} - 2 B D * D E = 2 A D^{2} - 2 D E^{2} + 2 B D^{2} + 2 D E^{2} = 2 A D^{2} + 2 B D^{2} = 2 (A D^{2} + B D^{2})$ which is the required result Regards

Prove that in any triangle the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median, which bisects the third side.