central angle theorem proof

(i) ( Theorem) From (i) and (ii), ..

Similarly it can be proved that Hence the theorem is prov..

Now in triangles, ABD and ABC, AB=AB (common side) (each angle is a right angle) AD=BC (opposite sides of parallelogram) BD=AC (corresponding parts of corresponding triangles) Hence the theorem is prov..

In triangles ABC and DBC, AB=DC (opposite sides of parallelogram) BC=BC (common side) AC=BD (given) (corresponding parts of corresponding triangles) But these angles are consecutive interior angles on the same side of transversal BC and AB||DC. By definition of rectangle, parall..

Let case I - when Gaussian surface is spherical in shape, a positive charge q be placed at the centre of an imaginary spherical surface of radius R as shown in the figure. By symmetry, the field due to the charge +q is radial and E is perpendicular to the sphere and is directed along the normal to ..

Suppose P is different from O, then in triangles OPM and OPN, OP=OP (common side) (given) ( given) (AAS congruency theorem) Hence, PM=PN ( CPCT) If P coincides with O, then it is equidistant from AB and CD, since they intersect at O P is equidistant from AB and ..

x.(x+y) = (x + 0)(x + y) 2(a) (Refer to Axion in previous topic) = x + (0.y) Axiom 4(a) = x + (y.0) Axiom 3(b) = x + 0 (Theorem 2) = x Axiom 2..

MN is a transversal. (Alternate angles) . . (2) (Alternate angles) From the figure, ( is a straight line and sum of the angles at M = 180 o ) From (1) and (2)..

( Given) (Vertically opposite angles) (corresponding angles) AB||CD ( corresponding angles axio..

( Linear pair) ( corresponding angles axiom) Similarly, we can prove that..