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 $OP=CM$ .

View step-by-step help

Home

Math Problems

Algebra 2

Write equations of cosine functions using properties

Full solution

Q. 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

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 $CD$ $0$ . The line $CD$ $1$ is perpendicular to $CD$ $2$ , which means that triangle $CD$ $3$ 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 $CD$ $2$ 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 $PON$ $1$ ).
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 $ON$ $0$ .
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 $\frac{AM}{PM} = \frac{ON}{PN}$ . But since $ON = PN$ (because triangle $PON$ is isosceles with $ON$ and $PN$ being radii), we have $PON$ $2$ .
Isosceles Right Triangle POM: If $\frac{AM}{PM} = 1$ , 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 $PO = OM$ .
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 $45$ $0$ (the radius of the circle) and triangle $POM$ is an isosceles right triangle, $45$ $2$ must also equal $45$ $3$ . Therefore, $45$ $4$ , which shows that $45$ $2$ equals $45$ $6$ .