 Hello and welcome to the session. In this session we discussed the following question that says construct a triangle ADC in which AD equal to 6 cm, BC equal to 7 cm and CA equal to 6.5 cm, find a point P equidistant from B and C and also equidistant from AD and BC. Let us consider some useful method to be used in the construction. Locals of a point equidistant. The straight line is the given. Let's proceed with the solution now, ADC in which AD is 6 cm, BC is 7 cm and CA is 6.5 cm. In the next step, we have 6 cm and in the next step, we have 6.5 cm. At this point of intersection of the two arcs, we point next step, we join the triangle ADC in which AD is of measure 6.5 cm and BC is of measure 7 cm. ADC is the required triangle ADC. Let's suppose to find a point P which is equidistant from the points B and C equidistant from the sides AD and BC. Equidistant from two given points is the right point P which is equidistant from the points B and C would be the right by sector of BC. That is the line joining the points B and C. The right by sector side BC, by sector of the side BC would be equidistant from the points B and C. Next we know that the locals of the points equidistant lines would be the by sector of the angles between the given intersecting straight lines point P which would be equidistant from the sides AD and BC also. So we will draw the angle by sector formed between the straight lines AD and BC, angle by sector, angle by sector of the angle ADC. Now as you can see that the right by sector of the side BC and the angle by sector of angle ADC intersect at also from the points B and BC. That point I should hope you have understood the solution of this question.