Answer:

Given triangle ABC with AC^2 = AB^2 + BC^2 ,and triangle LMN constructed such that LM = AB , MN = BC , and angle LMN = 90^circ .

Now, let's prove that triangle ABC is right-angled at B

In right-angled triangles, the Pythagorean theorem states that the square of the hypotenuse (in this case, AC is equal to the sum of the squares of the other two sides AB and BC .

AC^2 = AB^2 + BC^2

This is exactly what is given in triangle ABC . Therefore, triangle ABC is a right-angled triangle, and the right angle is at B .

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Answer:

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Step-by-step explanation:

To prove that triangle ABC is right-angled at B, we can use the given information and apply the Pythagorean theorem.

Given:

AC² = AB² + BC²

LM = AB

MN = BC

LM ⊥ MN (perpendicular)

To prove:

∠ABC is a right angle (ABC is right-angled at B)

Proof:

From the given information, we know that LM = AB and MN = BC. Since LM ⊥ MN, we can conclude that triangle LMB and triangle MNB are right-angled triangles.

In triangle LMB:

By Pythagoras theorem,

LM² + MB² = LB²

Substituting LM = AB, we have:

AB² + MB² = LB² -- (1)

In triangle MNB:

By Pythagoras theorem,

MN² + NB² = MB²

Substituting MN = BC and NB = AC, we have:

BC² + AC² = MB² -- (2)

From equation (1) and equation (2), we can rewrite equation (2) as:

AB² + BC² + AC² = LB²

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

AB² + BC² + AB² + BC² = LB²

Simplifying the equation:

2AB² + 2BC² = LB²

Dividing both sides by 2:

AB² + BC² = (LB²)/2 -- (3)

Comparing equation (3) with the Pythagorean theorem (AC² = AB² + BC²), we can see that both equations are the same. Therefore, LB = AC.

Since LB = AC, triangle ABC satisfies the condition of the Pythagorean theorem, and we can conclude that ∠ABC is a right angle.

Hence, triangle ABC is right-angled at B (ABC is right-angled at B).