rmo aakash lecture 7 1 oct 2026
Q1. S1. Draw the ||gram first. That will make it easier to draw the rest of the diagram. Let O be center of circumcircle of BCD BOD is 2.theta. OB = OD = r Given AB = AD = x. => AO is perpendicular bisector of BD. Now we can show BOD and BAD are congruent triangles. BD common. Both are isosceles so AO is angle bisector of 2.theta for both. Angle ABD = ADB = OBD = ODB = 90 - theta By ASA congruent. So corresponding sides are equal. => r = x. BODA is Rhombus. Now CP is also r cause ABCP is a ||gram. Since BODA is Rhombus AB || OD. Also ABCP is ||gram so AB || CP => AB || OD || CP Also AB = CP = r = OD OD = CP and OD || CP => ODPC is a ||gram. => DP = OC = r H.P. that CP = DP.