Question

Please solve this one

In the Triangle ABC, AC^2 = AB^2 + BC^2. If CE and AF are the medians of the Triangle. Show that, AF^2 = 1/4(5AC^2 - 4CE^2)

​​

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• A ∆ ABC in which AF and EC are median
• AE=EB and BF=FC

${\sf{\green{\underline{\large{To\:find}}}}}$

• AF²=1/4(5AC²-4CE²)

${\sf{\pink{\underline{\Large{Explanation}}}}}$

Fom big ∆ABC

• right angled at b

From Pythagoras Theorm

AB²+BC²=AC²

In ∆ ABF

AB²+BF²=AF²______(2)

In ∆ EBC

BE²+BC²=EC²

=>(1/4AB²)+4BF²= EC² ______(3)

Now joing equation 2 and 3

AB²+BF²=AF²

(1/4AB²)+4BF²= EC²

________________

(5/4 AB²)+5BF²=AF²+EC²

Now we conclude

_________________________

(5/4 AB²)+(5/4Bc²)=AF²+EC²

=>5/4(AB²+BC²)=AF²+EC²

=>5/4(AC²) =AF²+EC²

=>5AC²=4AF²+4EC²

=>5AC²-4EC³=4AF²

=>1/4(5AC²-4EC²)=AF²

__________________________

Henec,proved