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Preeti Dabral 2 years, 1 month ago
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.
Posted by Umang Upadhyay 2 years, 1 month ago
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Preeti Dabral 2 years, 1 month ago
(6q+1)(6q+2)(6q+3)
in all 6q is common
then 6q is out
=6q(1+2+3)
=6q(6)
=36q/6
=6q
Yes, this is divisible by 6.
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