Two perpendicular diameters $AB$ and $CD$ are considered on a circle. Let $M$ be a point on $BC$. Prove that $AN=MN+MB$.

Question

Problem

Two perpendicular diameters $AB$ and $CD$ are considered on a circle with centre $O$. Let $M$ be a point on the small arc $BC$. The parallel drawn through $O$ to $MD$ intersects $AM$ in $N$. Prove that $AN=MN+MB$.

The drawing

My idea

As you can see I noted $\angle OMD=x$.

Because $NO$ is parallel with $MD$, we can say that $\angle OMD= \angle NOM=x$

$MO=OD=R => \angle OMD= \angle ODM= x$

We can simply realise that $ \angle COM= \angle ODM*2 => \angle NOC=x$

Because $NO$ bisects $\angle MOP$ we can apply the bisector theorem:

$\frac{PN}{MN}=\frac{PO}{MO}$

I also thought that triangles $NOA$ and $MBD$ are similar.

I don't know what to do forward. Hope one of you can help me! Thank you!

Answer 1

Make $NE=NM$, draw $CE$ through to $F$, join $FA$, $FB$, $NC$, and draw $ON$ through to $G$.

Since $\angle DMC=90^o$ and $OG\parallel DM$, and hence by SAS$$\triangle NGC\cong\triangle NGM$$then$$NM=NC=NE$$whence $C$ lies on the circle about diameter $ME$, and $\angle MCE=90^o$.

And since$$\angle BMD=\angle DMA=\angle AMC=45^o$$ then $MCE$ is an isosceles right triangle similar to triangle $EAF$, and$$AF=AE$$and in inscribed rectangle $MAFB$,$$AF=MB$$Therefore$$AE=MB$$and$$AE+EN=AN=MB+MN$$

Answer 2

Hint :

Drop perpendiculars from center to point $K$ on $BM$ and point $L$ on $AM$. What can you say about the shapes $\square \,OKML$ and $\triangle \,OLN \,? $

Calculate the lengths $AN$, $BM+MN$ and show they are equal.