In the given figure, LM and …

To show that the perimeter of AABL is constant, we need to prove that AB + BA + AL + BL is constant.

First, let us label the angles in the figure as follows:

∠NLM = ∠LNM = α (since LM and LN are tangents from an external point, they are equal in measure)
∠MAB = β (angle between the tangent AB and the chord MC)
∠NBA = β (angle between the tangent AB and the chord NC)
∠ALN = γ (angle between the tangent LN and the chord MN)
∠BLM = γ (angle between the tangent LM and the chord MN)

Using the fact that the sum of angles in a triangle is 180 degrees, we can write:

∠LAM = 180 - α - β
∠LBN = 180 - α - β

Now, let us consider the perimeter of AABL:

AB + BA + AL + BL

Using the fact that angles in the same segment of a circle are equal, we can write:

∠MCN = ∠MAB + ∠NBA = 2β

Using the fact that the opposite angles in a cyclic quadrilateral add up to 180 degrees, we can write:

∠ALB = 180 - ∠MCN = 180 - 2β

Now, let us consider the triangles LAM and LBN. Using the fact that angles in a triangle add up to 180 degrees, we can write:

∠LAM + ∠ALN + ∠ALM = 180
∠LBN + ∠BLM + ∠BLN = 180

Substituting the values we have calculated for ∠LAM and ∠LBN, and using the fact that ∠ALN = ∠BLM = γ, we get:

(180 - α - β) + γ + ∠ALM = 180
(180 - α - β) + γ + ∠BLN = 180

Simplifying, we get:

∠ALM = α + β - γ
∠BLN = α + β - γ

Finally, let us consider the perimeter of AABL again:

AB + BA + AL + BL = AB + BA + 2AL sin γ + 2BL sin γ

Using the fact that AL = AM and BL = BN, and using the sine rule in triangles LAM and LBN, we get:

AB + BA + 2AM sin γ + 2BN sin γ = AB + BA + 2LM sin (α + β - γ) + 2LN sin (α + β - γ)

Using the fact that LM = LN (since they are tangents from an external point), we get:

AB + BA + 2LM sin (α + β - γ) + 2LM sin (α + β - γ)

Simplifying, we get:

AB + BA + 4LM sin (α + β - γ)

Using the fact that sin (α + β - γ) = sin (180 - γ) = sin γ, we get:

AB + BA + 4LM sin γ

Substituting the value of LM (which is equal to LN), we get:

AB + BA + 4LN sin γ

But we know that sin γ is a constant, since it is determined by the position of point L and the circle MN. Therefore, the perimeter of AABL is constant, since it does not depend on the position of point C on the arc MN.

Hence, we have proved that the perimeter of AABL is constant.