Q3 of 15 Page 82

A chord and the diameter through one of its ends are drawn in a circle. A chord of the same inclination is drawn on the other side of the diameter.


Prove that the chords are of the same length.


EF is the diameter of the circle.


Chord EG and EH both have same inclination. GF and HF are joined.


We have to prove that, EG = EH.


In ΔGEF and ΔHEF we have,


∠GEF = ∠HEF [∵ both have same inclination]


EF is the common side.


∠EGF = ∠EHF = 90° [∵ both are angle inscribed in a semi circle]


∴∆GEF≅∆HEF [AAS congruency]


∴ EG = EH [similar sides of congruent triangles]


More from this chapter

All 15 →
1

Prove that the line joining the centres of two intersecting circles is the perpendicular bisector of the line joining the points of intersection.

 2

The picture on the right shows two circles centred on the same point and a line intersecting them.


Prove that the parts of the line between the circles on either side are equal.

 4

Prove that the angle made by two equal chords drawn from a point on the circle is bisected by the diameter through that point.

 5

Draw a square and a circle through all four vertices. Draw diameters parallel to the sides of the square and draw a polygon joining the end points of these diameters and the vertices of the square.


Prove that this polygon is a regular octagon.