 Hello and welcome to the session. Let us discuss the following question. It says line L is a bisector of an angle A and B is any point on L. BP and BQ are perpendicular from B to the arms of angle A. Show that triangle ABB is congruent to triangle AQB and BP is equal to BQ or B is equidistant from the arms of angle A. Now to solve this question we will be using ASA congruence criteria that is angle side angle congruence criteria. Let us understand what is ASA criteria. It says that if we have two triangles say ABC and BEF then the triangles are congruent by ASA criteria if two angles and one included side of one triangle is equal to the two angles and one included side of the other triangle. So this knowledge will work as K idea. This now proceed on with the solution. We are given that L is the bisector of angle A and we are also given that BQ is perpendicular to AQ and BP is perpendicular to AP. Now we have to prove the triangle APB is congruent to triangle AQB. So now in triangle APB and triangle AQB angle AQB is equal to angle APB because they are right angles. Also the side AB is common to both the triangles that is AB is equal to AB since it is a common side. QAB is equal to angle BAP. This is because L bisects angle A at angle QAB is equal to angle BAP and angle AQB is equal to angle APB. Now to use ASA congruence criteria we need to prove that angle ABQ is equal to angle ABB. Which is true. Angle ABQ is equal to angle ABP because we have proved that two angles of the two triangles are equal. Therefore third angle is also equal. So we have proved that two angles at one included side of the triangle ABP and AQB are equal. So by ASA criteria triangle APB is congruent to triangle AQB and also we know that corresponding parts of congruent triangles are congruent. So BQ is equal to BP by CPCTC. Hence we have proved that triangle APB is congruent to triangle AQB and BQ is equal to BP. Hence the result is proved. So this completes the question. Hope you enjoyed the session. Lüben Teklif.