 Welcome to the session. I am Aashira and I am going to help you with the following question that says in figure 6.33 PQ and RS are two mirrors placed parallel to each other and incident ray AB Strikes the mirror PQ at B The reflected ray moves along the path BC. This is the reflected ray and strikes the mirror RS at C And again reflects back along CD Proves that AB is parallel to CD. Let us now begin with the solution. First, let us draw normals B, M and RS respectively. Write the construction draw normals is perpendicular on parallel to each other. Now perpendicular to the line RS therefore these two lines are B, M parallel to CD. And now since B, M is parallel to C, M once versus which intersects them implies angle, this angle is equal to CB Just this angle. Let us name the angle's flexion Incidence is equal to the angle of reflection that is V is equal to X Y is equal to equal to the angle of reflection Equal to Y on multiplying we have 2x is equal to 2y Equal to Y plus Y Now since X is equal to V. So replacing 1x by V We have V plus X is equal to and since Y is equal to U replacing 1 Y by U We have on the right-hand side U plus Y Now V plus X is angle ABC U plus Y is angle DCB Let this be number one. Thus we have angle ACB is equal to angle BCD implies line AB is parallel to CD Since if a transversal intersects two lines. So this implies Line AB is parallel to the solution. Hope you enjoyed it. Take care and bye for now