 Hello and welcome to the session. Let us understand the following question today. In figure 6.54, O is a point in the interior of a triangle ABC where OD is perpendicular to BC, OE is perpendicular to AC and OF is perpendicular to AV. Show that AF square plus VD square plus CE square is equal to AE square plus CD square plus VF square. Now here we have the figure 6.54. ABC is a triangle where OD is perpendicular to BC, OF is perpendicular to AB and OE is perpendicular to AC and we have joined OA, OB and OC. Now in right triangles, triangle ODB and triangle ODC. By Pythagoras theorem, we have OB square is equal to OD square plus VD square and OC square is equal to OD square plus CD square. Let us name it as 1 and 2. Now subtracting 2 from 1, we get OB square minus OC square is equal to OD square plus VD square minus OD square minus CD square. Here we see that this gets cancelled with this. So we get OB square minus OC square is equal to BD square minus CD square. Let us name it as number 3. Now similarly, OC square minus OA square is equal to CE square minus AE square. Let us name it as number 4. And OA square minus OB square is equal to AF square minus VF square. Let us name it as number 5. Now adding 3, 4 and 5, we get OB square minus OC square plus OC square minus OA square plus OA square minus OB square which is equal to BD square minus CD square plus CE square minus AE square plus AF square minus VF square. Now here we see that OB square gets cancelled with minus OB square, OC square gets cancelled with minus OC square and OA square gets cancelled with minus OA square. So we are left with which implies taking minus CD square minus AE square minus BA square to the left side, we get CD square plus AE square plus VF square is equal to BD square plus CE square plus AF square. Or it can be written as AF square plus VD square plus CE square is equal to AE square plus BF square plus CD square. This is our required result. Hence proved. I hope you understood the question. Bye and have a nice day.