In right-angled triangle ABC in which $∠ C = 90$ , If D is the mid-point of BC, prove that ${A B}^{2} = 4 {A D}^{2} - 3 {A C}^{2}$ .

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Solution

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In $△ A C D$

$A D^{2} = A C^{2} + C D^{2}$ .....(1)

In $△ A C B$

$A B^{2} - A C^{2} = B C^{2}$ .....(2)

As $C D = D B = \frac{B C}{2}$ ...... (3)

Substituting (3) in (1)
$A D^{2} = A C^{2} + {\frac{B C}{4}}^{2}$

$4 A D^{2} - 4 A C^{2} = B C^{2}$ ....... (4)

Subtracting (2) from (4)

$4 A D^{2} - 4 A C^{2} - (A B^{2} - A C^{2}) = B C^{2} - B C^{2}$

$4 A D^{2} - 4 A C^{2} - A B^{2} + A C^{2} = 0$

$4 A D^{2} - 3 A C^{2} = A B^{2}$

Hence proved.

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Similar questions

ABC is a right angled triangle at C. If D is the mid point of BC,prove that AB²=4AD²-3AC²

Δ

B C^{2} = 4 (A D^{2} - A C^{2})

∠ B = 90^{\circ}

B C

D

E

8 {A E}^{2} = 3 {A C}^{2} + 5 {A D}^{2}

△ A B C, ∠ A C B = 90^{\circ}

{A B}^{2}

{4 A D}^{2} - {3 A C}^{2}

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