seg ab and seg ac of triangle abc are the diameters of the circle intersecting side bc in point d such that d divides side bc in the ratio 3:1 then prove that ab2 = ac2+1/2bc2
Answers
Answer:
GIVEN: A triangle ABC, AD perpendicular to
BC. & BD: DC = 3:1
TO PROVE: 2( AB? - AC? ) = BC?
Since BD: DC = 3:1
SO. BD = 3x & DC = X
In right triangle ADB,
AB? = BD? + AD?
=> AB? = 9x? + AD2 (1)
In right triangle ADC
AC? = DC? + AD?
=> AC? = x? + AD? .(2)
Eq(1) - Eq(2)
=> AB? - AC? = 9x2 - x2 + AD? - AD?
=> 2( AB? - AC? ) = 2* 8x?
=> 2( AB? - AC? ) = 16 x? =(4x)2 = BC2
Step-by-step explanation:
pls mark brainliest
Given : Seg AB and seg AC of ∆ABC are the diameters of the circle intersecting side BC in point D such that D divides side BC
in the ratio 3 : 1
To Find : prove that
AB² = AC² + (1/2)BC²
Solution:
AB is a diameter
Hence ΔABD is right angle triangle
=> AD² = AB² - BD²
AC is a diameter
Hence ΔACD is right angle triangle
=> AD² = AC² -CD²
Equate
AB² - BD² = AC² -CD²
=> AB² = AC² + BD² -CD²
=> AB² = AC² + (BD+ CD)(BD - CD)
BD+ CD = BC
D divides side BC in the ratio 3 : 1
=> BD = 3BC/4 and CD = BC/4
BD - CD = 2BC/4 = BC/2
AB² = AC² + (BC)(BC)/2
=> AB² = AC² + (1/2)BC²
QED
Hence proved
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