If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
From the question we know the following:
(i) AB and CD are 2 chords which are intersecting at point E.
(ii) PQ is the diameter of the circle.
(iii) AB = CD.
Now, we will have to prove that BEQ = CEQ
For this, the following construction has to be done:
Construction:
Draw two perpendiculars are drawn as OM ⊥ AB and ON ⊥ D. Now, join OE.
The constructed diagram will look as follows:
Now, consider the triangles ΔOEM and ΔOEN.
(i) OM = ON [Since the equal chords are always equidistant from the centre]
(ii) OE = OE [It is the common side]
(iii) OME = ONE [These are the perpendiculars]
So, by RHS congruency criterion, ΔOEM ΔOEN.
Hence, by CPCT rule, MEO = NEO
∴ BEQ = CEQ (Hence proved).