 Hello and welcome to the session. Let's work out the following problem. It says three forces P, Q and R, F along O, A, O, V and O, C, where O is the orthocenter of the triangle ABC. If the forces are in equilibrium, prove that P is to Q is to R is equal to A is to W is to C. So let's now move on to the solution. We are given that the three forces P, Q and R act along O, A, O, V and O, C of the triangle ABC where O is the orthocenter of the triangle. Now since O is the orthocenter that means AD is perpendicular to VC, CF is perpendicular to AB and VE is perpendicular to AC. Now angle VOC is equal to angle FOE is equal to 180 degrees minus angle A. This is because in quadrilateral AFOE. Angle F is 90 degrees, angle E is 90 degrees. The sum of these angles is 180 degrees and angle FOE is equal to 360 degrees minus angle F minus angle E minus angle A. As we know that sum of all four angles of this quadrilateral is 360 degrees. So angle FOE is equal to 360 degrees minus 90 minus 90 minus angle A that means it is equal to 360 minus 180 minus angle A which is equal to 180 minus angle A and this is because angle AFO is equal to angle AEO is equal to 90 degrees. Similarly, angle AOB is equal to angle BOE is equal to 180 degrees minus angle C and similarly angle AOC is equal to angle DOF is equal to 180 degrees minus angle B. Now we are given that these three forces are in equilibrium. So since forces P, Q and R are in equilibrium therefore by Lemmy's theorem we have P upon angle sine of angle BOC is equal to Q upon sine of angle AOC is equal to R upon sine of angle AOB. As we know that Lemmy's theorem says if we have three forces which are in equilibrium acting at one point then each force is proportional to the sine of the angle between the other two forces. So P is proportional to the sine of angle between Q and R which is BOC and similarly we have Q upon sine of angle AOC and it is equal to R upon sine of angle AOB. Now sine of BOC is 180 degrees minus angle A. So this is P upon sine 180 minus angle A is equal to Q upon sine of 180 degrees minus angle B is equal to R upon sine of AOB which is 180 minus C this is again equal to P upon sine of angle A Q upon sine B is equal to R upon sine C as we know that sine of 180 minus theta is sine theta. Now also by sine formula we have A upon sine A is equal to B upon sine B is equal to C upon sine C where small a b c are the sides of the triangle a b c. Let's name this as one and this as two. Now from one and two we have P upon A is equal to Q upon B is equal to R upon C so we have P is to Q is to R is equal to A is to B is to C and this is what we have to prove. That's the question and the session why for now take care have a good day.