Given : Circle C (P,r) and circle C (Q, r’) intersects each other at the points A and B.
To prove: points p and Q lies on the perpendicular bisector of common chord AB.
Construction: Join point P and Q to mid-point M of chord AB.
Proof: AB is chord of circle C (P, r) and PM is bisector of chord AB.
Therefore, PM ⊥AB
[∵ The line drawn through the center of a circle to bisect a chord is perpendicular to the chord.]
Hence, ∠PMA = 90°
Similarly, AB is chord of circle C (Q, r) and QM is bisector of chord AB.
Therefore, QM ⊥AB
[∵ The line drawn through the center of a circle to bisect a chord is perpendicular to the chord.]
Hence, ∠QMA = 90
Now, ∠PMA + ∠QMA = 90° + 90° = 180°
Since, ∠PMA and ∠QMA are forming liner pair. So PMQ is a straight line.
Hence, points P and Q lies on the perpendicular bisector of common chord AB.

I/Steps of construction
(i) Draw a ray AB at the point A.
(ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at F.
(iv) Similarly, taking E as centre and with the same radius, draw an arc which intersect at E.
(v)Taking E and F as as center, draw arcs with equal radius (more than half of EF). which intersect at H.
(vi) Draw a ray AH which intersects the main arc at D.
(vii) Taking F and D as centre, draw arcs with equal radius (more than half of FD), which intersect at G.
(viii) Draw a ray AG. ∠GAB is the required angle of 75°.

II/Steps of construction.
(i) Draw a ray AB at the point A.
(II) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as center and with the same radius, Draw an arc which intersect the previous arc at D.
(iv) Similarly, taking E as centre and with the equal radius, draw an arc which intersect at G.
(v) Taking D and G as centre, draw arcs with same radius (more than half of DG), which intersect at F.
(vi) Draw a ray AF which intersects the main arc at E.
(vii) Taking E and G as centre, draw arcs with equal radius (more than half of EG), which intersect at H.
(viii) Draw a ray AH. ∠HAB is the required angle of 105°.

III/Steps of construction.
(i) Draw a ray AD at the point A.
(II) Taking B as centre and a convenient radius, draw an arc which intersect AD at C.
(III) Taking C as centre and with the same radius, draw an arc which intersect the previous arc at P.
(iv) Similarly, Taking P as centre and with the equal radius, draw an arc which intersect at F.
(v) Taking P and F as centre, draw arcs with same radius (more than half of PF), which intersect at H.
(vi) Draw a ray BH from the point B.
(vii) Taking B as center, draw an arc taking some radius, which intersects AB at E and BH at Q.
(viii) Taking E and Q as centre, draw arcs with equal radius (more than half of EQ), which intersect at G.
(ix) Draw a ray BG. ∠GBD is the required angle of 135°.

Steps of construction
(i) Draw a line segment BC = 7 cm.
(ii) Using ruler and compass, draw an angle ∠CBX = 75°.
(iii) Taking B as centre and 13 cm as radius, mark an arc on BX, which intersects at D.
(iv) Jion CD and draw a perpendicular bisector (PQ) of CD, which intersects BD at A.
(v) join AC.
(vi) Triangle ABC is the required triangle.
Justification
Point A lies on the perpendicular bisector of DC. So, AD = AC
Here, AB = BD – AD
⇒ AB = BD – AC [∵ AD = AC]
⇒ AB + AC = BD

Steps of construction
(i) Draw a line segment QR = 6 cm.
(ii) At point Q, Using ruler and compass, draw an angle ∠RQX = 60°. Produce XQ to K.
(iii) Taking Q as centre and 2 cm as radius, draw an arc which intersects QK at S.
(iv) Jion SR and draw a perpendicular bisector (MN) of SR, which intersects QX at point P.
(v) join PR. Triangle PQR is the required triangle.
Justification
Point P lies on the perpendicular bisector of SR. So, PS = PR
Here, QS = PS – PQ
⇒ QS = PR – AC [∵ PS = PR]

Steps of construction
(i) Draw a line segment AB = 12 cm.
(ii) At point A, Using ruler and compass, draw an angle ∠BAX = 90°.
(iii) Taking A as centre and 18 cm as radius, draw an arc which intersect AX at D.
(iv) Join B to D. Draw a perpendicular bisector (MN) of BD which intersects AD at C.
(v) join B to C. Triangle ABC is the required triangle.
Justification
Point C lies on the perpendicular bisector of BD.
So, BC = CD
Here, AD = AC + CD
⇒ AD = AC + BC [∵ BC = CD]