 Hello and welcome to the session. Let us understand the foreign question today. In figure 6.54, O is a point in the interior of a triangle ABC. OD is perpendicular to BC. OE is perpendicular to AC and OF is perpendicular to AB. Show that OA squared plus OB squared plus OC squared minus OD squared minus OE squared minus OF squared is equal to AF squared plus BD squared plus CE squared. Now let us see the figure. Here we have the figure 6.54. In which O is the interior of a triangle ABC, OD is perpendicular to BC and OF is perpendicular to AB. Now let us first join OA, OB and OC. Here we have joined OB, OA and OC. Now triangle OFA is a right angle triangle since OF is perpendicular to AB. Therefore, by Pythagoras theorem, OA squared is equal to AF squared plus OF squared. Let us name this as 1. Similarly, in triangles OBD and OEC, we can write AB squared is equal to BD squared plus OD squared. Let us name it as 2. And OC squared is equal to CE squared plus OE squared. Let us name it as 3. Now adding 1, 2 and 3, we get it implies OA squared plus OB squared plus OC squared is equal to AF squared plus OF squared plus BD squared plus OD squared plus CE squared plus OE squared. That is, which implies OA squared plus OB squared plus OC squared taking my OD, OF and OE on left hand side we get minus OD squared minus OE squared minus OF squared is equal to AF squared plus BD squared plus CE squared. And this is our required result, hence proved. I hope you have answered the question. Bye and have a nice day.