NCERT Class 10 Mathematics

TRIANGLES 149

or, BL^2 =

2

AC AB 2

2

✁ ✂

✄✆ ☎✝ (L is the mid-point of AC)

or, BL^2 =

2

AC AB 2

4

✞

or, 4 BL^2 =AC^2 + 4 AB^2 (2)

From ✟ CMA,

CM^2 =AC^2 + AM^2

or, CM^2 =AC^2 +

2

AB

2

✁

✄ ☎

✆ ✝

(M is the mid-point of AB)

or, CM^2 =AC^2 +

AB^2

4

or 4 CM^2 =4 AC^2 + AB^2 (3)

Adding (2) and (3), we have

4 (BL^2 + CM^2 ) = 5 (AC^2 + AB^2 )

i.e., 4 (BL^2 + CM^2 ) = 5 BC^2 [From (1)]

Example 14 : O is any point inside a

rectangle ABCD (see Fig. 6.52). Prove that
OB^2 + OD^2 = OA^2 + OC^2.

Solution :

Through O, draw PQ || BC so that P lies on

AB and Q lies on DC.

Now, PQ || BC

Therefore, PQ ✠ AB and PQ ✠ DC (✡ B = 90° and ✡ C = 90°)

So, ✡ BPQ = 90° and ✡ CQP = 90°

Therefore, BPQC and APQD are both rectangles.

Now, from ✟ OPB,

OB^2 =BP^2 + OP^2 (1)

Fig. 6.52