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