Justify why AB in Fig. is the perpendicular bisector. Class 7
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Justify why AB in Fig. is the perpendicular bisector. Class 7
Question 1.
Justify why AB in Fig. is the perpendicular bisector. Class 7

Solution:
Join A and B.
Meeting XY at 0.

In ∆AXB and ∆AYB, here
AX =AY (Equal ropes)
BX = BY (Equal ropes)
AB = AB.
So, ∆AXB ≅ ∆AYB (SSS)
Hence, ∠XAO = ∠YAO (CPCT) ............... (1)
Now, in ∆AXO and ∆AYO,
AX = AY, AO = AO
and ∠XAO = ∠YAO [From (1)]
So, ∆AXO ≅ ∆AYO (SAS)
Hence, XO = YO
and ∠AOX = ∠AOY (CPCT) ...(2)
= 90°
From (2), AB is the perpendicular bisector of XY.
Question 2.
Can you think of different methods to construct a 90° angle at a given point on a line using a rope? Class 7
Solution:
Yes.