 Welcome friends once again to this new session on gems of geometry and today we are going to discuss another theorem. So the theorem says the locus of a point equidistant from two intersecting lines is the angle y sector of the angles between the intersecting lines. So this is the theorem which we are going to prove. Before that let's understand what's the meaning of locus. So locus word comes from Latin origin. When Latin locus means place, location and what does locus mean in mathematics especially in geometry is nothing but it is a set of all points. It could be a line, a line segment, a curve or a surface. So it's basically a set of points such that the point satisfy certain conditions, so any set of point which satisfy one particular condition will be called a locus of that particular point. So if you see if there is a point which has which is equidistant from a given point on a plane of the paper and today we know that that is called a circle. So once again so let's say there is a point O and we are trying to find out the set of points P which are you know which are equidistant from O that means the distance O P stays constant is nothing but if you trace it out you will find that this is nothing but a circle. Isn't it if you draw a circle then all the points on the circle are equidistant from a particular point called the center of the circle. So hence this particular path or this particular trace will be called the locus of point P. So it will go and you know form a circle like that. So don't go by the you know the drawing. So basically I will you know if you take a circle so if you take a circle so this is a circle let's say so hence every point every point on the circumference is equidistant from the center so hence it is called the locus of this point which is equidistant from a particular given point O right. So hence circle becomes a locus of a point P. So now we are going to discuss in this case the locus of a point again we have to trace the path of a point such that it is equidistant from two intersecting lines so in this case if you see L and M are two intersecting lines and we have a point P which is equidistant from both the lines. So what does equidistant means if you drop a perpendicular let's say like these two perpendicular if you drop let's say this is a and this is B then we know that OA is equal to OB that's what we know and now we have to find out the locus of this point P locus means if I have another P here if I in such a way that if I drop perpendicular like that they are again equal then if I join all such point P's all such point P's right for example here also there is a point like that and they are equidistant it is equidistant from this thing. So the theorem says that if you join all of them like that you will see that this line which you will get after joining all of them you'll see that the line you will get after joining all of them will be nothing but the angle bisector of the angle between the lines L and M. So let's try and prove this so I hope that you understood the theorem. Now let's understand let's try and prove this theorem now how to prove this theorem. So let's say this point was P this point is P I'm rewriting it this is P okay and we have dropped two perpendicular P a and P b so now we know that you know P a is equal to P b P a is equal to P b it's you know because they are B point P is equidistant from both L and M correct now what you join so let's say given I'm writing given P a is equal to P b and we have constructed P a perpendicular to P a perpendicular to L and P b perpendicular to M correct this is what we have done and now what we need to approve that we need to prove that to prove to prove to prove what so we are we are saying that a point P lies on the angle bisector of OAB so we hence we have to prove that angle a OP is equal to angle B OP this is what we need to prove correct so hence we have already joined OP now if you say in triangles in triangle OPA OPA and triangle OPB OPB what do we know P a is equal to P b why it's given the point P is equidistant correct now angle PAO is equal to angle P a P sorry POA POA angle P angle P angle PO sorry PAO it was correct actually PAO is equal to P b O why both are equal to 90 degrees because PA was perpendicular to L and PV was perpendicular to M right and OP is equal to OP common common side correct common side so hence what do we infer we infer that therefore triangle OPA OPA is congruent to triangle OPB OPB by which rule RHS congruence congruence criteria correct so if they are congruent that means this implies what do we know now angle POA it will be equal to angle P OB and why it is CPCT right that means if a point P is equidistant from both the lines then it is also lying on the angle by sector of the you know the angle which is formed by intersection of the two given lines so hence proved hence hence points points which are which are equidistant equidistant equidistant from equidistant from two intersecting lines two intersecting lines also lie on the angle by sector angle by sector of the of the angles of the angle formed formed by the lines okay so this is what we learned so so this is what we learned right so this is nothing but in other way we can say that the locus of all such point P which are equidistant from the two given line is nothing but a angle by sector of the given angle