Q-9 Two congruent circles intersect each other at point P and Q. Through P any line segment APB is drawn so that A, B lie on the two circles. Prove that QA=QB.

Solution:

Given,

Two congruent circles intersecting each other at point P and Q.

PQ is a common chord of these circles.

Since PQ is common chord

Angle subtended by arcs of equal lengths on congruence circles must be equal, therefore,

Therefore in isosceles triangle AQB,

Q-10 In any triangle PQR, if the angle bisector of and perpendicular bisector of QR intersect, prove that they intersect on the circumcircle of the circumcircle of triangle PQR.

Solution:

Given,

PQR is a triangle inscribed in a circle with centre at O, A is a point on the circle

PA is the internal bisector of and B is the mid-point of QR

Prove:

BA is the right bisector of QR

In , since both the arcs QA and AR subtend equal angle at the circumference (.

Therefore in triangle QAR,

(Given)

(common line)

Therefore, by SSS criterion of congruence,

Hence, (corresponding parts of congruent triangles)