if a b c d are in proportion then prove that a^2+ab+b^2 upon a^2-ab+b^2 =c^2+cd+d^2 upon c^2-cd+d^2
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8
Answer:
Step-by-step explanation:
a2 + c2), (ab + cd) and (b2 + d2) are in continued proportion.
⇒ (a2 + c2) : (ab + cd) = (ab + cd) : (b2 + d2)
⇒ (a2 + c2) (b2 + d2) = (ab + cd) (ab + cd)
⇒ a2b2 + a2d2 + c2b2 + c2d2 = a2b2 + 2abcd + c2d2
⇒ a2d2 + c2b2 – 2abcd = 0
⇒ (ad – cb)2 = 0
⇒ ad – cb = 0
⇒ ad = bc
⇒ a/b = c/d
⇒ a, b, c and d are in proportion.
Answered by
5
Answer:
a2 + c2), (ab + cd) and (b2 + d2) are in continued proportion.
⇒ (a2 + c2) : (ab + cd) = (ab + cd) : (b2 + d2)
⇒ (a2 + c2) (b2 + d2) = (ab + cd) (ab + cd)
⇒ a2b2 + a2d2 + c2b2 + c2d2 = a2b2 + 2abcd + c2d2
⇒ a2d2 + c2b2 – 2abcd = 0
⇒ (ad – cb)2 = 0
⇒ ad – cb = 0
⇒ ad = bc
⇒ a/b = c/d
⇒ a, b, c and d are in proportion
Step-by-step explanation:
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