Question

In a traingle right angled at B,AD and CE are medians then prove that 5AC2= 4(AD2*CE2)

Answer 1

Hey there !!


➡ Given :-

→ A ∆ABC in which AD and CE are medians and $\angle$B = 90° .

➡ To prove :-

→ 5AC² = 4( AD² + CE² ) .

➡ Proof :-

In ∆ABC, $\angle$B = 90° .

•°• AC² = AB² + BC² .......(1) .
[ By Pythagoras theorem ] .

In ∆ABD, $\angle$B = 90° .

•°• AD² = BD² + AB² .
[ By Pythagoras theorem ] .

=> AD² = ( ½BC )² + AB² .

=> AD² = ¼BC² + AB² .

=> 4AD² = BC² + 4AB² ..............(2).

In ∆BEC, $\angle$B = 90° .

•°• CE² = BE² + BC² .

=> CE² = ( ½AB )² + BC² .

=> CE² = ¼AB² + BC² .

=> 4CE² = AB² + 4BC²................(3) .

▶ On adding equation (2) and (3), we get

=> 4AD² + 4CE² = BC² + 4AB² + AB² + 4BC² .

=> 4( AD² + CE² ) = 5BC² + 5AB² .

=> 5( BC² + CE² ) = 4( AD² + CE² ) [ Using (1) ] .

•°• 5AC² = 4( AD² + CE² ) .


✔✔ Hence , it is proved ✅✅.


THANKS


#BeBrainly.

Answer 2

here is your answer OK ☺☺☺☺☺☺


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