rmo aakash lecture 7 1 oct 2026

 Q1.

Let $ABCD$ be a convex quadrilateral with $\angle BAD = 2\angle BCD$ and $AB = AD$. Let $P$ be a point such that $ABCP$ is a parallelogram. Prove that $CP = DP$.

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.

Q2.

The ratio of the median $AM$ of a triangle $ABC$ to the side $BC$ equals $\sqrt{3}:2$. The points on the sides of $ABC$ dividing these sides into 3 equal parts are marked. Prove that some 4 of these 6 points are concyclic.
S2.
Simple way to remember Appollonius theorem.

Step 1.

Step 2.
How to decide which 4 points are to be chosen?

Step 3.

From BPT(basic proportionality theorem also known as Thales proportionality theorem)
A1C2 || AB => Angle A1C2C  = Angle A (BAC).

Now if we prove theta1  = Angle A, are we done?
Yes because that would imply theta1 +theta2 = 180 and H.P.

using appollonius:

then
A1C1 = sqrt(3).a
B1C1 = 
B1A1 = 

isosceles trapezium is always concyclic?
what about ||gram?