Question

in triangle abc , angle b = 90° and d is the mid point of bc. prove that bc^2= 4(ad^2-ab^2)

Answer 1

Given,
→∆ABC where B=90°

→D is mod point of BC so CD=BD

To Prove

→BC^2=4(AD^2-AB^2)

PROOF:--

In ∆ABD by pythagorus theorem we get,

AD^2=AB^2+BD^2

SO
➡️BD^2= AD^2-AB^2. (1)

NOW ,

BC= BD+CD

BC= 2BD (AS D IS MID POINT)

SQUAREING BOTH SIDES

BC^2= 4BD^2

BY EQUATION 1 WE GOT BD^2 = AD^2-AB^2

Putting this value

➡️BC^2= 4(AD^2-AB^2)

HENCE PROVED

Answer 2

Answer: Hello

Step-by-step explanation: