B Why are BD and FE perpendicular to the curve BF C Verify t

Solution

Ans.

(B) G is the center of the curve ABF. The points B and F on curve ABF are close to each other. Similarly G, D and E on curve CDGE also close to each other. From given figure, it is clear that they are so close that tangent drawn on one point can be considered as the tangent at the other point also (on both the curves).

Hence, Tangent HB is tangent at F also.

We know, GF and GB are radius of the circle which can be made by curve ABF.

thus, from geometrical theorem, \"Radius is always perpendicular to the tangent.\"

GF and GB are perpendicular to the tangent HB. Since the distance between the points B and F is very small and points are on curve ABF. We can consider the curve between these two points as a straight line. And hence, GF and GB are perpendicular to curve BF.

(C) Line BO is parallel to line MN. Again, triangle MGN and triangle BGO has two sides as the part of the radius of the curve ABF. So these sides are in proportion as

GM/BG = GN/GO (since the distance between points B and F is very small there is no large distance between F and O). Also, triangle MGN and triangle BGO has an angle G in common. Thus, these triangles are similar triangles by SOS (SIde-Angle-Side) Rule.

Using the property of SOS Rule,

For triangle MGN and triangle BGO,

BG/GM = BO/MN

Hence Proved.