State and prove converse of Pythagoras theorem.?

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In a triangle  if the square of one side is equal to the sum of the squares of the other two sides then the angle opposite to the first side is a right angle ...

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Answer

Converse of the Pythagorean Theorem:

Statement: If the square of the length of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

Proof:

To prove the converse of the Pythagorean theorem, we assume that we have a triangle ABC, where AB, BC, and AC are the three sides of the triangle. We are given that the square of the length of one side, say AB, is equal to the sum of the squares of the other two sides, BC and AC.

Step 1: Assume that AB² = BC² + AC²

Step 2: We need to prove that triangle ABC is a right triangle, i.e., ∠BAC = 90 degrees.

Proof by contradiction:

Assume that triangle ABC is not a right triangle, i.e., ∠BAC ≠ 90 degrees.

Step 3: Let ∠BAC = θ, where θ is any angle other than 90 degrees.

Step 4: Since θ ≠ 90 degrees, we can draw a perpendicular line from point A to side BC, meeting at point D.

Step 5: Let AD be the perpendicular line and BD = x and DC = y.

Step 6: Using trigonometry, we can write AB, BC, and AC in terms of x, y, and θ.

AB = BD/sinθ = x/sinθ
BC = DC/cosθ = y/cosθ
AC = BD/tanθ = x/tanθ

Step 7: Now, we can find AB², BC², and AC² using the above expressions.

AB² = (x/sinθ)² = x²/sin²θ
BC² = (y/cosθ)² = y²/cos²θ
AC² = (x/tanθ)² = x²/tan²θ

Step 8: Adding BC² and AC², we get:

BC² + AC² = y²/cos²θ + x²/tan²θ

Step 9: Rearranging the terms, we have:

BC² + AC² = y²/cos²θ + x²/tan²θ
= [(y²cos²θ + x²sin²θ)/cos²θtan²θ]

Step 10: Simplifying further, we have:

BC² + AC² = [(y²cos²θ + x²sin²θ)/(cos²θsin²θ)]
= [(y² + x²sin²θ/cos²θ)/(sin²θ/cos²θ)]
= [(y² + x²tan²θ)/(sin²θ/cos²θ)]
= (y² + x²tan²θ)(cos²θ/sin²θ)
= (y²cos²θ + x²sin²θ)/sin²θ
= AB²/sin²θ

Step 11: Since AB² = BC² + AC² (given), we have:

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