# 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) OA^{2} + OB^{2} + OC^{2} - OD^{2} - OE^{2} - OF^{2} = AF^{2} + BD^{2} + CE^{2},

(ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}

**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.

Using Pythagoras theorem, in ΔOAF,

OA^{2} = AF^{2} + OF^{2} [Since, ∠OFA = 90°]............... (1)

Similarly, In △OBD

OB^{2} = BD^{2} + OD^{2} [Since ∠ODA = 90°].............. (2)

In △OCE

OC^{2} = CE^{2} + OE^{2} [Since, ∠OEC = 90°]………………(3)

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

OA^{2} + OB^{2} + OC^{2} = AF^{2} + OF^{2} + BD^{2} + OD^{2} + CE^{2} + OE^{2}

OA^{2} + OB^{2} + OC^{2} - OD^{2} - OE^{2} - OF^{2} = AF^{2} + BD^{2} + CE^{2}............. (4)

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

(OA^{2} - OE^{2}) + (OB^{2} - OF^{2}) + (OC^{2} - OD^{2}) = AF^{2} + BD^{2} + CE^{2}

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

Using pythagoras theorem we have,

AE^{2} = OA^{2} - OE^{2} .............. (5)

BF^{2} = OB^{2} - OF^{2} .............. (6)

CD^{2} = OC^{2} - OD^{2} ............... (7)

AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2 }[ 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 OA^{2} + OB^{2} + OC^{2} - OD^{2} - OE^{2} - OF^{2} = AF^{2} + BD^{2} + CE^{2 }and AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.

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