Let $C_1,C_2$ be two circles such that the center of $C_1$ is on the circumference of $C_2$. Let $C_1,C_2$ intersect each other at points $M,N$. Let $A,B$ be two points on the circumference of $C_1$ such that $AB$ is the diameter of it. Let lines $AM,BN$ meet $C_2$ for the second time at $A',B'$, respectively. Prove that $A'B'=r_1$ where $r_1$ is the radius of $C_1$.
Problem
Source:
Tags: geometry proposed, geometry
AK1024
23.09.2010 15:47
Sorry but I cannot provide a diagram.
Let $\angle OB'N =\angle MA'O= \angle MB'O = \angle MNO = \angle OMN = x$.
We have $\angle MON=180-2x \implies \angle MAN=x \implies \angle MBB'=90-x $. Hence $\angle B'MB=90-x$ since $%Error. "ANGLE" is a bad command.
BB'M=2x$. Since $AB$ is the diameter of $C_1$, $\angle AMB =\angle A'MB = 90$ so that $\angle A'MB' =x\implies \angle AOB'=x$. Thus we have $\triangle A'PB'$ and $\triangle MOP$ is congruent since $\triangle A'PM$ is isosceles so that $A'B'=MO=r_1$.
Bars
23.09.2010 18:56
(O - center C1) Angle ONB'=OBN=AMN=A'B'N, qed.
Mahdi_Mashayekhi
22.12.2021 19:07
Let O1 be center of C1. ∠O1MA = ∠BAM = ∠BNM = ∠B'O1M ---> B'O1 || MA' ---> A'B' = MO1 = r1