Question

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

Answer 1

(i) Applying Pythagoras theorem in ΔAOF, we obtain

OA² = OF² + AF²

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.

From above result

AF² + BD² + EC² = (OA² - OE²) + (OC² - OD²) + (OB² - OF²)

∴ AF² + BD² + CE2 = AE² + CD² + BF².