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
then prove: AB2 = AC2 +
1
2
BC2
Answers
here is the answer
hope it helps you
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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