Every Theorem on Circle with Proofs.| Theorem on Circles.| Class 9 |NCERT.

In this video, we will understand various theorems related to the circles and their proof. So let's continue with the first one. Equal codes obtain equal angles at the center. Consider a circle with center point O. Let AB and CD be the two equal codes. Join the endpoints of each code with center O, making angle AOB and COD respectively at the center. Triangles so formed are congruent as side AO is equal to side CO, which are the radii of circle. Similarly, side BO is equal to side DO are also the radii thus are equal. Side AB and side CD are equal as given codes are equal. Therefore, by SSS criteria, triangle AOB is congruent to triangle COD. By corresponding parts of congruent triangle, angle AOB is equal to angle COD. Hence, the theorem is proved. The converse of the theorem is also true. That is, if the angles subtended by the two codes at the center are equal, then the length of the codes are also equal. For the proof, only the condition of congruency is changed. As central angle A1 is equal to central angle A2, side AO is equal to side CO, which are the radii of the circle. Similarly, side BO is equal to DO, which are also the radii and thus are equal. By SAS rule, triangle AOB is congruent to triangle COD. By corresponding parts of congruent triangle, we conclude side AB is equal to side CD. Hence, the theorem is proved. Moving on to the next theorem. The theorem states that any line perpendicular from center to the code bisects the code. Consider a circle with center O. Let AB be the code and ON be the perpendicular on it, making angle ONA and angle ONB each 90 degree. Join OA and OB representing radius of the circle. Triangle ONA and triangle ONB can be proved congruent at side NO is common to both the side AO is equal to side BO each radius. Angle ANO is equal to angle BNO each 90 degree. Therefore, by RHS congruency, triangle ANO is congruent to triangle BNO. Hence, by corresponding parts of congruent triangle, we conclude side AN is equal to side NB. Hence, the theorem is proved. Moving on to the next theorem. The theorem states that the two equal codes of a circle are equidistant from the center. Consider a circle with center O. Let AB and CD be the two equal codes. Join AO and OC as both are radius. Construct perpendicular OX on code AB and OY on code CD, bisecting the codes. Therefore, AB can be expressed as twice of AX and CD can be expressed as twice of CY. As codes are equal, we can conclude AX is equal to CY. Now in triangle AOX and triangle COY, side AO is equal to side CO as both are radius. Side AX is equal to side CY. Angle AXO is equal to angle CYO each 90 degree. From construction, by SAS rule, triangle AOX is congruent to triangle COY. By CPCT, we conclude OX is equal to OY. Hence, the theorem is proved. Moving on to the next theorem. Theorem states that the angle subtended by the arc at the center is double the angle subtended by the same arc at any point on the remaining part of the circle. Consider a circle with center O and points A and B on the circle acting as minor arc. This minor arc subtends angle AOB at the center. Let P be any point on the remaining part of the circle. Here the remaining part can also be called the major arc of the circle. Join point P to point A and point B. To prove angle AOB as twice of angle APB, we need to construct a line from point P such that it passes through point O. From figure, angle APB can be written as the sum of angle 1 and angle 2 while angle AOB can be written as the sum of angle 3 and angle 4. Now consider the isosceles triangle AOP as sides AO is equal to PO which are the radii and thus are equal. Since base angles of isosceles triangle are equal, so angle PAO is equal to angle APO. Further, in a triangle, the exterior angle is equal to the sum of interior opposite angles. Therefore, angle 3 can be expressed as twice of angle 1. Similarly, in isosceles triangle BOP, angle 4 can be written as twice of angle 2. As angle AOB is the sum of angle 3 and angle 4, so on adding we get angle AOB as twice of angle APB. Hence the theorem is proved. Moving on to the next theorem. The theorem states that angle in a semicircle is 90 degree. Consider a circle with center O. Let A and B be the points on the circle such that AB acts as the diameter resulting in the formation of two semicircles. Angle AOB measuring 180 degree is formed at the center point O. Now consider a point P on one of the semicircle. Join end points of the diameter with the point P. Angle APB so formed will always be equal to 90 degree. As angle subtended by the arc at the center is double the angle subtended by the same arc at any point on the remaining part of the circle. Thus angle APB is equal to half of 180 degree. Hence the theorem is proved. Moving on to the next theorem. The theorem states that only one unique circle can pass through three non-collinear points. Consider ABC as three non-collinear points. Join point A and B. Now construct perpendicular bisector of AB. As perpendicular bisector of the code always passes through the center of the circle, so to find the center of the circle on the perpendicular bisector line of AB, join point B to C and construct perpendicular bisector of it. The point where these two bisectors meet will be the center of the circle and is denoted by point O. Hence the theorem is proved. Moving on to the next theorem. The theorem states that angles formed in the same segment of a circle are equal. Consider a circle with center O. Join points A and B of a circle resulting in the formation of minor segment and major segment. Now consider any point P on the circle such that angle APB equal to X. Join endpoints of the line segment AB with center O, resulting in the formation of angle AOB equal to 2X. As angle subtended by the arc at the center is double the angle subtended by the same arc at any point on the remaining part of the circle. Consider another point Q on the same side of the segment such that angle AQB is equal to Y. Now angle at the center with reference to angle AQB is 2Y. It is clear from the figure that angle AOB with reference to point P and point Q is same. So 2 times angle X is equal to 2 times angle Y. By cancellation law, angle X is equal to angle Y. Hence the theorem is proved. This was it for the video. See you in the next one. Thank you for watching.