In Fig. 6.54, O is a point in the interior of a triangleABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

(i)

(ii)

Join OA, OB, and OC

(i) Applying Pythagoras theorem in ΔAOF, we obtain

OA² = OF² + AF²

Similarly, in ΔBOD,

OB² = OD² + BD²

Similarly, in ΔCOE,

OC² = OE² + EC²

Adding these equations, we get

OA² + OB² + OC² = OF² + AF² + OD² + BD² + OE² + EC²

OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + EC²

(ii) From the above given result,

AF² + BD² + EC² = (AO² – OE²) + (OC² – OD²) + (OB² – OF²)

Therefore,

AF² + BD² + EC² = AE² + CD² + BF²