Question

AD is the altitude from A in the ∆ABC and DB:CD = 3:1 . prove that 2(AB²-AC²) = BC²​

Answer 1

Answer:

GIVEN : A triangle ABC , AD perpendicular to BC. & BD : DC = 3:1

TO PROVE : 2( AB² - AC² ) = BC²

Since BD: DC = 3:1

SO. BD = 3x & DC = x

In right triangle ADB,

AB² = BD² + AD²

=> AB² = 9x² + AD² …………..(1)

In right triangle ADC

AC² = DC² + AD²

=> AC² = x² + AD² ………….(2)

Eq(1) - Eq(2)

=> AB² - AC² = 9x² - x² + AD² - AD²

=> 2( AB² - AC² ) = 2 * 8x²

=> 2( AB² - AC² ) = 16 x² =(4x)² = BC²

Hence Proved