Let AB and CD be two perpendicular diameters of a circle with centre O. Consider point M on the diameter AB, different from A and B. The line CM cuts the circle again at N. The tangent at N to the circle and perpendicular at M and AM intersect at P. Show that OP=CM.

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Identify Properties and Points: Let's start by identifying the properties of the circle and the points involved. We know that AB and CD are diameters of the circle and are perpendicular to each other, which means that they intersect at the center O of the circle. Since M is on diameter AB, OM is a radius of the circle. The line CM intersects the circle at N, which means CN is also a radius of the circle. Since a tangent at a point on a circle is perpendicular to the radius at that point, PN is perpendicular to ON. The line PM is perpendicular to AM, which means that triangle PAM is a right triangle with the right angle at M.Radius of Circle: Since CM is a line segment from the center of the circle to a point on the circumference, it is a radius of the circle. Therefore, CM equals the radius of the circle. Let's denote the radius of the circle as r. So, CM = r.Right Triangle OCM: Since AB and CD are diameters and are perpendicular, triangle OCM is a right triangle with the right angle at C. Since OM is a radius, OM also equals r. Therefore, in triangle OCM, we have OC = r and OM = r, which means triangle OCM is an isosceles right triangle. This implies that angle OCM is 45 degrees.Similar Triangles PAM and PON: In triangle PAM, since PM is perpendicular to AM, and we know that ON is perpendicular to PN (because PN is a tangent to the circle at N), triangles PAM and PON are similar by AA (Angle-Angle) similarity (both have a right angle and share angle P).Proportional Sides: In similar triangles, corresponding sides are proportional. Since OM and ON are both radii of the circle, OM = ON = r. Therefore, the ratio of OP to CM is the same as the ratio of PN to AM. Since CM = r, we need to show that OP also equals r to prove that OP = CM.Isosceles Right Triangle PAM: Since triangles PAM and PON are similar, and OM = ON, we can say that AM is to PM as ON is to PN. This means that AM/PM = ON/PN. But since ON = PN (because triangle PON is isosceles with ON and PN being radii), we have AM/PM = 1.Isosceles Right Triangle POM: If AM/PM = 1, this means that AM = PM. Since M is on AB and PM is perpendicular to AM, triangle PAM is also an isosceles right triangle, which means angle PAM is 45 degrees.Conclusion: Since angle PAM is 45 degrees and angle OCM is 45 degrees, and both triangles PAM and OCM are isosceles right triangles, triangle POM, which is formed by combining triangles PAM and OCM, is also an isosceles right triangle with PO = OM.Since OM = r (the radius of the circle) and triangle POM is an isosceles right triangle, OP must also equal r. Therefore, OP = OM = CM = r, which shows that OP equals CM.

Let $AB$ and $CD$ be two perpendicular diameters of a circle with centre $O$ . Consider point $M$ on the diameter $AB$ , different from $A$ and $B$ . The line $CM$ cuts the circle again at $N$ . The tangent at $N$ to the circle and perpendicular at $M$ and $AM$ intersect at $P$ . Show that $OP=CM$ .

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