Problem Statement:
Two circles of the same radius pass through the center of each other and intersect each other at points P and N. A line PM is drawn to intersect one of the circles at points L and M. We need to prove that triangle LMN is an equilateral triangle.

Proof:

Step 1: Establishing the Given Information
- Let O1 and O2 be the centers of the two circles.
- Since the circles pass through the center of each other, O1P and O2N are radii of the respective circles.
- Let the radius of the circles be r.
- The line PM intersects the circle at points L and M.

Step 2: Observations
- Since O1P and O2N are radii of their respective circles, they are equal in length, i.e., O1P = O2N = r.
- Since the two circles are of the same radius, O1O2 = 2r.
- The line PM intersects the circle at points L and M, implying that LM is a chord of the circle.

Step 3: Proving the Triangle is Equilateral
- Since O1O2 = 2r and O1P = r, we can conclude that O1P is half the length of O1O2. Therefore, angle O1PN is a right angle.
- Similarly, since O2N = r and O1O2 = 2r, we can conclude that O2N is half the length of O1O2. Therefore, angle O2NP is a right angle.
- Since angle O1PN is a right angle and angle O2NP is a right angle, we can conclude that angle PNO1 and angle PNO2 are both 90 degrees.
- Since angle PNO1 and angle PNO2 are both 90 degrees, we can conclude that PNO1O2 is a rectangle.
- In a rectangle, all angles are 90 degrees and opposite sides are equal in length.
- Therefore, O1P = O2N = NO1 = NO2 = r.
- Since LM is a chord of the circle, it is equidistant from the center of the circle (O2N) at point N.
- Therefore, LN = NM.
- Similarly, since LM is a chord of the circle, it is equidistant from the center of the circle (O1P) at point P.
- Therefore, LP = PM.
- Since LN = NM and LP = PM, we can conclude that LMN is an equilateral triangle.

Conclusion:
In the given scenario, we have proved that triangle LMN is an equilateral triangle.

Can anyone give me solution of this question