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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

Solution:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

(i) In △ABC

OD ⊥ BC, OE ⊥ AC and OF ⊥ AB

OA, OB and OC joined as shown below.

In Figure 6.54, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.

Using Pythagoras theorem, in ΔOAF,

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)]]

☛ Check: NCERT Solutions for Class 10 Maths Chapter 6

Video Solution:

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

Summary:

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 + CE²and AF² + BD² + CE² = AE² + CD² + BF².

☛ Related Questions:

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

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

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

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