 Hi and how are you all today? My name is Priyanka and the question says EBC is a triangle locate a point in the interior of triangle EBC which is equidistant from all the vertices of triangle EBC. Now we should know that the point of concurrency of the perpendicular bisectors of the sides of the triangle is called the circumcentre. It is equidistant from the vertices of triangle and that is given to us as ABC. And the knowledge of this important information is the key idea of this question. So what we are going to do is we will be finding out the perpendicular bisectors of all the sides of the triangle and the point where they need that is the point of concurrency will be a point which is equidistant from all the vertices. Let us proceed on with our solution. Now I will be making a very rough figure over here and telling you all the instructions and that will be your steps of construction. But you will be in your copies making it very neatly with the help of a scale and a compass. The first step is to draw a triangle ABC. Let us have a triangle over here. I am again telling you that this is a rough figure that I will be making to make your understanding more clear with all the steps of construction. Now the second step is to construct the perpendicular bisectors that you know how to construct and you learnt in your seventh and eighth classes of sides let's say A, C and the point where these perpendicular bisectors will meet that will be your point of concurrency. Now for AB taking a centre as B and radius more than half of AB draw an arc here and here. Similarly taking A as a centre and radius more than half of AB draw another two arcs intersecting the previous arcs like this and then join these two points. Similarly draw a perpendicular bisector for C also in the same manner let them intersect at point O. Right now the third perpendicular bisector that will be of BC will also pass through O. So O is the circumcentre of this triangle ABC. Now what you need to do next is you need to join A to O, B to O and C to O. By measurement you will find that the length of AO is equal to OB is equal to OC and this is the point O that is equidistant the vertices C of triangle AB. So O lies on the perpendicular bisectors of AB AC and hence it is equidistant from end points of AB and AC. So this was our required point to be found. This completes our session. Hope you enjoyed it. Make your diagram very neat and clean. This was just a rough diagram to make your understandings of these steps more clear. I hope you enjoyed. Take care. Bye for now.