$AC = AB$ , radius of the $\widehat{BC}$
$BC = AB$ , radius of the $\widehat{AB}$
Thus, $AC = BC$
$AC = AB$ , radius of the $\widehat{BC}$
The placement of point $X$ is not clear to me.I didn't understand the $2$nd line of it.Durjoy Sarkar wrote: ↑Mon Mar 05, 2018 10:06 pmradius of two big circle are same. let X be the point where little circle is tangent.
it is well known the center of little circle, tangent point are lies on $OB$.
$MO=OX$
$BX= BO+OX=BO+MO $
You proved $BX=OB+OM$, not $AB=OB+OM$Durjoy Sarkar wrote: ↑Mon Mar 05, 2018 10:06 pmradius of two big circle are same. let X be the point where little circle is tangent.
it is well known the center of little circle, tangent point are lies on $OB$.
$MO=OX$
$BX= BO+OX=BO+MO $
Look carefully and you will notice that $BX$ and $AB$ are radii of the same circle centered $B$Tasnood wrote: ↑Wed Mar 07, 2018 10:07 amYou proved $BX=OB+OM$, not $AB=OB+OM$Durjoy Sarkar wrote: ↑Mon Mar 05, 2018 10:06 pmradius of two big circle are same. let X be the point where little circle is tangent.
it is well known the center of little circle, tangent point are lies on $OB$.
$MO=OX$
$BX= BO+OX=BO+MO $