Consider an isosceles triangle ABC with sides BC = 30, CA = AB = 20. Let D be the foot of the perpendicular from A to BC, and let M be the midpoint of AD. Let PQ be $a$ chord of the circumcircle of triangle ABC, such that M lies on PQ is parallel to BC. The length of PQ is:

Step 1: Eq. of PQ : $y$ =

Step 2: Perpendicular bisector of AC:

Step 3: 5 7 −1 15

Step 4: $y$ − = x−

Step 5: 2 5 7 − 0 2

Step 6:

Step 7: 5 7 15 15

Step 8: => $y$ − = x−

Step 9: 2 5 7 2

Step 10: Centre  intersection of perpendicular bisector

Step 11: −5 −5 7

Step 12:  0,  0,

Step 13: 7 7

Step 14: => Eqn of circumcircle

Step 15: 2 2

Step 16: 5 7 5 7

Step 17: ( $x$ − 0 ) + $y$ +

Step 18: = 5 7 +

Step 19: 7 7

Step 20: 5 64 25 \times 64

Step 21: x2 + $y$ + = 25 \times 7 \times =

Step 22: 7 49 7

Step 23: Intersection with PQ : $y$ =

Step 24: 25 \times 64 5 7 5 25 \times 64

Step 25: 452 ( )

Step 26: 7

Step 27: 25 \times 64 \times 4 − 452 52 64 \times 4 − 81 52

Step 28: x2 = = 2 = 2 \times 25

Step 29: 28 2 7 2

Step 30: 25 25 25

Step 31: => x= => distance = − − = 25

Step 32: 2 2 2