**Proof:**

To prove the equation acosA bcosB ccosC = 2asinBsinC, we will use trigonometric identities and properties.

**Step 1: Rewrite the equation**

Let's rewrite the equation in terms of sine functions:

acosA bcosB ccosC = 2asinBsinC

**Step 2: Convert cos to sin using the identity**

We can use the identity cosθ = √(1 - sin^2θ) to convert the cosine functions to sine functions:

asinA √(1 - sin^2B) √(1 - sin^2C) = 2asinBsinC

**Step 3: Simplify the equation**

Now, let's simplify the equation further:

asinA √((1 - sin^2B)(1 - sin^2C)) = 2asinBsinC

**Step 4: Expand the product**

Using the identity (a - b)(a + b) = a^2 - b^2, we can expand the product in the square root:

asinA √(1 - sin^2B - sin^2C + sin^2Bsin^2C) = 2asinBsinC

**Step 5: Simplify the square root**

We can simplify the square root by combining the terms inside it:

asinA √(1 - (sin^2B + sin^2C - sin^2Bsin^2C)) = 2asinBsinC

**Step 6: Rearrange the terms**

Let's rearrange the terms to isolate the sine functions on one side:

asinA = 2asinBsinC / √(1 - (sin^2B + sin^2C - sin^2Bsin^2C))

**Step 7: Use Pythagorean identity**

We can use the Pythagorean identity sin^2θ + cos^2θ = 1 to simplify the equation further:

asinA = 2asinBsinC / √(1 - (1 - cos^2B)(1 - cos^2C) - sin^2Bsin^2C)

**Step 8: Simplify the equation**

Now, let's simplify the equation by expanding the terms inside the square root:

asinA = 2asinBsinC / √(1 - (1 - cos^2B - cos^2C + cos^2Bcos^2C) - sin^2Bsin^2C)

**Step 9: Combine like terms**

We can combine the like terms inside the square root:

asinA = 2asinBsinC / √(cos^2B + cos^2C - cos^2Bcos^2C - sin^2Bsin^2C)

**Step 10: Use the identity**

Finally, we can use the identity cos^2θ = 1 - sin^2θ to simplify the equation:

asinA = 2asinBsinC / √((1 - sin^2B) + (1 - sin^2C) - (1 - sin^2B)(1 - sin^2C) - sin^2Bsin^2C)

**Step 11: Simplify the equation further**

Let's simplify the equation by expanding and combining the terms

By the sine rule, we have
(a / sin A) = (b / sin A) = (c / sin A) = k ⇒ a = ksin A, b = ksin B and c = ksin C.
LHS = acos A + bcos B + ccosC
= ksin A cos A + ksin B cos B + ksin C cos C
= (1/2) k [ sin 2A sin 2B sin 2C ]
= (1/2) k [ 2sin (A+B) cos (A-B) + 2sin C cos C ]
= k [ sin (A+B) cos (A-B) + 2sin C cos C ]
= k [ sin (π-C) cos (A-B) + sin C cos C ]
= k [ sin C cos (A-B) + sin c cos C ]
= k [ sin C { cos (A-B) + cos C } ]
= k [ sin C { cos (A-B) + cos (π - (A+B)) } ]
= k [ sin C { cos (A-B) - cos (A+B) } ]
= k [ sin C ( 2 sin A sin B ) ]
= k 2 sin A sin B sin C
= 2(ksin A) sin B sin C
= 2a sin B sin C