In Figure 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
OA, OB and OC joined as shown below.
Using Pythagoras theorem, in ΔOAF,
OA2 = AF2 + OF2 [Since, ∠OFA = 90°]............... (1)
Similarly, In △OBD
OB2 = BD2 + OD2 [Since ∠ODA = 90°].............. (2)
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)]]
In Figure 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
NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.5 Question 8
In the above figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Hence proved that OA2 + OB2 + OC2 - OD2 - OE2 - OF2 = AF2 + BD2 + CE2 and AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
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