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Answers
❁❁ Refer To Image First .. ❁❁
||✪✪ Given ✪✪|| :- AD is the Median and AE is perpendicular to BC.
|| ✰✰ Construction ✰✰ || :- Draw AE that is perpendicular to BC.
❁❁ Proof ❁❁ :-
In right triangles AEB and AEC, using Pythagoras theorem
→ AB² + AC² = ( BE² + AE² ) + ( EC² + AE² )
→ AB² + AC² = 2AE² + (BD - ED)² + (ED + DC)²
→ AB² + AC² = 2AE² + 2ED² + BD² + DC²
Now, since AD is median, so, BD = DC .
→ AB² + AC² = 2AE² + 2ED² + 2BD²
→ AB² + AC² =2(AE² + ED² + BD²)
→ AB² + AC² = 2(AD² + BD²)
now, again since AD is median , BD = 1/2 BC.
→ AB² + AC² = 2AD² + 2*(BC/2)²
→ AB² + AC² = 2AD² + (BC²/2)
✪✪ Hence Proved ✪✪
Given :- AD is the Median and AE is perpendicular to BC.
Construction:- Draw AE that is perpendicular to BC.
Proof :- Refer To Image ...
In right triangles AEB and AEC, using Pythagoras theorem
=> AB² + AC² = ( BE² + AE² ) + ( EC² + AE² )
=> AB² + AC² = 2AE² + (BD - ED)² + (ED + DC)²
=> AB² + AC² = 2AE² + 2ED² + BD² + DC²
As AD is median, so, BD = DC
→ AB² + AC² = 2AE² + 2ED² + 2BD²
→ AB² + AC² =2(AE² + ED² + BD²)
→ AB² + AC² = 2(AD² + BD²)
given that, AD is median , So, it divide BC in two Equal parts .
→ AB² + AC² = 2AD² + 2*(BC/2)²
→ AB² + AC² = 2AD² + (BC²/2) (Hence Proved)
