Question

someone please ans it correctly​

Answer 1

Answer:

if i know then i will surely say you

but i don't know anything

Step-by-step explanation:

mark brainliest

Answer 2

$\sf\huge\bold{Answer:}$

$\fbox{option (C) is correct statement}$

Given :

△ABC

 AD, BE and CF are medians .

To find :

Relation between sides of triangle and medians.

Solution :

In △ABC

 AD, BE and CF are medians to sides BC, AC and AB respectively.

So,

$\sf⇒BD=DC=½BC$

$\sf⇒AE=EC=½AC$

$\sf⇒AF=FB=½AB$

In △ABC

 AD, BE and CF are medians to sides BC, AC and AB respectively.

So,

$\sf⇒BD=DC=½BC$

$\sf⇒AE=EC=½AC$

$\sf⇒AF=FB=½AB$

By Pythagoras theorem in △AFC

$:⟶AC²=CF²+AF²$

$:⟶AC²=CF²+(½AB)²$

$:⟶AC²=CF²+¼AB²$

$:⟶4AC²=4CF²+AB²... (I)$

By Pythagoras theorem in △BCE

$:⟶BC²=BE²+EC²$

$:⟶BC²=BE²+(½AC)²$

$:⟶BC²=BE²+¼AC²$

$:⟶4BC²=4BE²+AC²... (II)$

By Pythagoras theorem in △ADB

$:⟶AB²=AD²+BD²$

$:⟶AB²=AD²+(½BC)²$

$:⟶AB²=AD²+½BC²$

$:⟶4AB²=4AD²+BC² ... (III)$

Adding (I),(II) and (III), we get

$\sf:⟶4AC²+4BC²+4AB²=4AD²+BC²+4BE²+AC²+4CF²+AB²$

$\sf:⟶3AC²+3BC²+3AB²=4AD²+4BE²+4CF²$

$\sf:⟶3(AC²+BC²+AB²)=4(AD²+BE²+CF²)$

$\sf\huge\purple{3(AB²+BC²+AC²)=4(AD²+BE²+CF²)}$