Altitudes AA1 and BB1 are drawn in acute △ABC. Prove that A1C·BC=B1C·AC
1 answer:
Answer:
See the attached figure which represents the problem.
As shown, AA₁ and BB₁ are the altitudes in acute △ABC.
△AA₁C is a right triangle at A₁
So, Cos x = adjacent/hypotenuse = A₁C/AC ⇒(1)
△BB₁C is a right triangle at B₁
So, Cos x = adjacent/hypotenuse = B₁C/BC ⇒(2)
From (1) and (2)
∴ A₁C/AC = B₁C/BC
using scissors method
∴ A₁C · BC = B₁C · AC
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