In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that (i) OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + CE2 (ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2 Eduinfy.com

NCERT Solution: Triangles

NCERT Exercise Exercise 6.5

In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that (i) OA2 + OB2 + OC2 − OD2 − OE2 − OF2 = AF2 + BD2 + CE2 (ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2

Join OA, OB, and OC

(i) Applying Pythagoras theorem in ΔAOF, we have

OA² = OF² + AF²

Similarly, in ΔBOD
OB² = OD² + BD²
Similarly, in ΔCOE
OC² = OE² + EC²
Adding these equations,
OA² + OB² + OC² = OF² + AF² + OD² + BD² + OE²+ EC²
OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE².

(ii) AF² + BD² + EC² = (OA² - OE²) + (OC² - OD²) + (OB² - OF²)
∴ AF² + BD² + CE² = AE² + CD² + BF².