In Fig. 6.54, 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

 

 

 

 

 

 

 

 

 

 

 

 

 

(i) In △ABC

OD ⊥ BC, OE ⊥ AC and OF ⊥ AB

OA² = AF² + OF² [Since, ∠OFA = 90°]............... (1)

Similarly, In △OBD

OB² = BD² + OD² [Since ∠ODA = 90°].............. (2)

In △OCE

OC² = CE² + OE² [Since, ∠OEC = 90°]………………(3)

Adding equations (1), (2) and (3)

OA² + OB² + OC² = AF² + OF² + BD² + OD² + CE² + OE²

OA² + OB² + OC² - OD² - OE² - OF² = AF² + BD² + CE²............. (4)

(ii) From equation (4) let's rearrange and group the terms.

(OA² - OE²) + (OB² - OF²) + (OC² - OD²) = AF² + BD² + CE²

Since, △OAE, △OFB and △ODC are right triangles

Using pythagoras theorem we have,

AE² = OA² - OE² .............. (5)

BF² = OB² - OF² .............. (6)

CD² = OC² - OD² ............... (7)

AF² + BD² + CE² = AE² + CD² + BF² [ From equations (5), (6) and (7)]]