Altitudes AA1 and BB1 are drawn in acute △ABC. Prove that A1C·BC=B1C·AC

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

898.1666667

step-by-step explanation:

5389/6 = 898.1666667

i cant answer this, theres no table. sorry man

step-by-step explanation: