The perpendicular AD on the base BC ofABCintersects BC in D such that BD=3DC.Prove that $2 A B^{2} = 2 A C^{2} + B C^{2}$

Open in App

Solution

$A B^{2} = A D^{2} + D B^{2}$
$O r, A B^{2} = A D^{2} + 9 C D^{2}$ (because $D B = 3 C D$ ) . . . . . (1)
$I n Δ A D C : A D^{2} = A C^{2} - C D^{2}$
Substituting the value of AD in equation (1), we get;
$A B^{2} = A C^{2} - C D^{2} + 9 C D^{2} = A C^{2} + 8 C D^{2}$
$O r, 2 A B^{2} = 2 A C^{2} + 16 C D^{2}$ . . . . . (2)
Since $B C = C D + 3 C D = 4 C D$
Hence, $B C^{2} = 16 C D^{2}$
Substituting the value from above equation in equation (2), we get;
$2 A B^{2} = 2 A C^{2} + B C^{2}$

Suggest Corrections

Similar questions

△

A D

B C

△ A B C

B C

D

D B = 3 C D

2 {A B}^{2} = 2 {A C}^{2} + {B C}^{2}

The perpendicular from A on side BC of a ΔABC intersect BC at D such that DB = 3 CD. Prove that 2 AB² = 2 AC² + BC²

Δ A B C

2 A B^{2} = 2 A C^{2} + B C^{2}

Δ A B C

2 A B^{2} = 2 A C^{2} + B C^{2}

Join BYJU'S Learning Program