give in triangle ABC, D is the midpoint of side BC and AE is perpendic

1.	give in triangle ABC, D is the midpoint of side BC and AE is perpendicular to BC. Proove that AB^2 + AC^2 = 2AD^2+ 2DC^2
Answer» Given: Since D is the MIDPOINT, BD = DC AE is PERPENDICULAR to BC To prove: AB² + AC² = 2AD² + 2DC² Solution: By Pythagoras theorem AB² = AE² + BE² ------ [Equation 1] AC² = AE² + CE² ------ [Equation 2] AD² = AE² + DE² ------ [Equation 3] We know that; CE = CD - DE Squaring on both sides, (CE)² = (CD - DE)² → CE² = CD² + DE² - 2CD. DE → CD² = CE² - DE² + 2CD. DE Adding Equation 1 and 2, AB² + AC² = AE² + AE² + BE² + CE² Now, BE² can be written as (BD + DE)² and CE² can be written as (CD - DE)² Taking the RHS part; *AE²* + *AE²* + BD² + DE² + 2BD. DE + CD² + DE² - 2CD. DE AE² and DE² can be GROUPED together. So it becomes 2AD². 2AD² + BD² + DC² + 2BD. DE - 2CD. DE Since BD = CD, 2AD² + DC² + DC² + 2BD. DE - 2BD. DE → 2AD² + DC² ∴ AB² + AC² = 2AD² + 2DC²

give in triangle ABC, D is the midpoint of side BC and AE is perpendicular to BC. Proove that AB^2 + AC^2 = 2AD^2+ 2DC^2

Answer»

Given:

To prove:

AB² + AC² = 2AD² + 2DC²

Solution:

By Pythagoras theorem

We know that;

CE = CD - DE

Squaring on both sides,

(CE)² = (CD - DE)²

→ CE² = CD² + DE² - 2CD. DE

→ CD² = CE² - DE² + 2CD. DE

Adding Equation 1 and 2,

AB² + AC² = AE² + AE² + BE² + CE²

Now,

BE² can be written as (BD + DE)² and

CE² can be written as (CD - DE)²

Taking the RHS part;

AE² + AE² + BD² + DE² + 2BD. DE + CD² + DE² - 2CD. DE

AE² and DE² can be GROUPED together.

So it becomes 2AD².

2AD² + BD² + DC² + 2BD. DE - 2CD. DE

Since BD = CD,

2AD² + DC² + DC² + 2BD. DE - 2BD. DE

→ 2AD² + DC²