Maths

TRIANGLES 149
or, BL 2 =
✄
AC
2
✁
☎
2
✂
AB
2
✆
✝
(L is the mid-point of AC)
or, BL 2 =
AC
4
✟
or, 4 BL 2 =AC 2 + 4 AB 2 (2)
From CMA,
CM 2 =AC 2 + AM 2
2
✞
AB
2
or, CM 2 =AC 2 +
✄
AB
2
✁
☎
2
(M is the mid-point of AB)
✆ ✝
or, CM 2 =AC 2 +
2
AB
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
Fig. 6.52
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)