Question

In a right angle ∆ABC, angle BAC=90°.segmentsAD,BE,CF are medians.prove that:2(AD SQUARE+BE SQUARE+CF SQUARE)=3BC SQUARE.

Answer 1

math]AE+EB=AB[/math]

2. [math]BD+DC=BC[/math]

3.[math] CF+FA=CA[/math]


adding 1 and 2 gives,

4. [math]AD + DC =  AB + BC[/math]

adding 2 and 3 gives,

5. [math]BF+FA =  BC + CA[/math]

adding 3 and 1 gives,

6. [math]CE+EB= CA + AB[/math]


Adding 4, 5 & 6 gives,

AD+BF+CE + DC+FA+EB = 2(AB+BC+CA)

AD+BF+CE + [math]\frac{1}{2}[/math](AB+BC+CA) = 2(AB+BC+CA)

(since D, E & F are the midpoints of AB, BC and CA)

but, from triangle law of vector addition we have,

[math]AB+BC+CA = 0[/math]

therefore,

[math]AD+BF+CE=0[/math]

Q.E.D

In the above proof I have used the notation that [math]AB[/math] represents [math]\vec{AB}[/math] just to avoid writing vector signs everywhere.