Question

If D, E, F are the mid-points of the
sides BC, CA, AB of a right triangle ABC
(rt. angled at A) respectively, prove
that 3(AB×AB+BC×BC+CA×CA)=4(AD×AD+BE×BE+CF×CF).​

Answer 1

Answer:

As BAE has a right angle at A, Pythagoras' Theorem gives

   4 BE² = 4 ( AB² + AE² ) = 4 ( AB² + (AC/2)² ) = 4 AB² + AC²

As CAF has a right angle at A, Pythagoras' Theorem gives

   4 CF² = 4 ( AC² + AF² ) = 4 ( AC² + (AB/2)² ) = 4 AC² + AB²

As ABC has a right angle at A, the triangle lies in a semicircle with centre at D and radius AD = BD = CD.  So...

   4 AD² = 4 BD² = 4 (BC/2)² = BC².

Adding these together now gives

   4 ( AD² + BE² + CF² )

=  BC² + 4 AB² + AC² + 4 AC² + AB²

=  5 AB² + 5 AC² + BC²

=  3 AB² + 3 AC² + 2 ( AB² + AC² ) + BC²

=  3 AB² + 3 AC² + 2 BC² + BC²               [ by Pythagoras' Theorem ]

=  3 ( AB² + AC² + BC² )

Hope this helps!

Answer 2

Step-by-step explanation:

the answer has been described