Question

In the figure given above, ABC is a right triangle right angled at C. If D is mid-point of BC, prove that AB^2 = 4AD^2 -3AC^2 ​

Answer 1

$\LARGE{ \underline{\underline{ \pink{ \bf{Required \: answer:}}}}}$

GiveN:

• ∆ABC is a right angled triangle, right angled at C.
• D is the midpoint of base BC.

To Prove:

• AB² = 4AD² - 3AC²

Step-wise-Step Explanation:

Refer to the attachment for the diagram with basic information and labellings.

Using Pythagoras theoram in ∆ACD:

⇒ AC² + DC² = AD²

⇒ AC² + (BC/2)² = AD²

⇒ AC² + BC² / 4 = AD²

Multiplying 4 with the above equation,

⇒ 4AC² + BC² = 4AD² ...............(1)

Now using Pythagoras theoram in ∆ACB:

⇒ AC² + BC² = AB² ..........(2)

Subtracting eq. (2) from eq. (1),

⇒ 4AC² + BC² - AC² - BC² = 4AD² - AB²

⇒ 3AC² = 4AD² - AB²

This can be written as:

⇒ AB² = 4AD² - 3AC²

And this is what we had to prove!!

Answer 2

Answer:

$\huge \orange{ \overbrace{ \red{\underbrace \color{pink} {\underbrace \color{lime}{\colorbox{black} { {\red \: {Answer}}}}}}}}$

$\huge { \boxed{ \underline{ \sf{ \orange{✯Given✧:-}}}}}$

• ∆ABC is a right angled triangle, right angled at C.

• D is the midpoint of base BC.

$\huge { \boxed{ \underline{ \sf{ \orange{✯To \: prove✧:-}}}}}$

• AB² = 4AD² - 3AC²

$\huge { \boxed{ \underline{ \sf{ \orange{✯Solution✧:-}}}}}$

• Using Pythagoras theoram in ∆ACD:

⇒ AC² + DC² = AD²

⇒ AC² + (BC/2)² = AD²

⇒ AC² + BC² / 4 = AD²

• Multiplying 4 with the above equation,

⇒ 4AC² + BC² = 4AD² ...............(1)

• Now using Pythagoras theoram in ∆ACB:

⇒ AC² + BC² = AB² ..........(2)

• Subtracting eq. (2) from eq. (1),

⇒ 4AC² + BC² - AC² - BC² = 4AD² - AB²

⇒ 3AC² = 4AD² - AB²

• It can be also written as:

⇒ AB² = 4AD² - 3AC²