Ch11 Q1
Q.1. To divide a line segment AB in the ratio 5 : 6 at a point P, a ray AX is drawn such that $\angle$BAX is an acute angle. Then a ray BY is drawn parallel to AX and the points $A_1, A_2, A_3, .... and B_1, B_2, B_3, ...$ are located at equal distances on rays AX and BY, respectively. Which points should be joined to obtain Point P? also write the names of the similar triangles so formed?

Hi, 
Since BAX is a acute angle so draw line AB of any length and then draw line AX as an acute angle forming XAB now it is given that 
AB is parallel to BY so take an radius of AM and mark an arc that cuts AX at some point R and then measure that arc and taking 
equal measure from point B mark an arc of radius BN = AM , down the line AB from point B
Now in a similar way from point N mark an arc of radius MR = NS that will cut the previous arc at B to the point S 
Join BS and extend it till Y so you will get AX parallel BY. 
Now taking equal distances draw points A₁, A₂,.....A₇ on line AX and B₁, B₂, ..B₇ on line BY 
Now we need the ratio 5:6 so take A₅ on AX and B₆ on BY join them and they will cut at P to AB so AP:PB = 5:6
 
Now similar triangle are AA₅P and BB₆P
​because angle A₅PA = angle B₆PB { vertically opposite angles }
and angle XAB = angle YBA 
so similar using AA similarity