duality theorem proof

Theorem 6(a) (Demorgan's Law)..

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..

When p(x) is divided by x-a, R = p(a) (by remainder theorem) p(x) = (x-a).q(x)+p(a) (Dividend = Divisor x quotient + Remainder Division Algorithm) But p(a) = 0 is given. Hence p(x) = (x-a).q(x) Conversely if x-a is a factor of p(x) then p(a)=0. p(x) = (x-a).q(x) + R If (x-a) is a facto..

The proof follows from theorem 2, P( f ) C = 1 - P( f ) = 1 - 1 = ..

Since the diagonals AC and BD are equal, ABCD is a rectangle - - -(i) (Diagonal property of rectangle) Since the diagonals are perpendicular to each other. ABCD is a rhombus. AB=AD - - -(ii) ABCD is a rectangle. (from i) With consecutive sides equal. (from ii) ABCD is a square. (by definition of a ..

By definition of complement of an element it is sufficient to prove xy + (x' + y') = 1 and xy(x' + y') = 0 xy + (x' + y') = (xy + x') + y' (Associativity of +) = (x + x')(y + x') + y' (Axiom 4b) = 1(y + x') + y' (Axiom 5) = y + x' + y' (2b) or 1 . x = x = y + y' + x' (Axiom 3a) = 1 + x' (Axiom 5) ..

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

Let AC and BD intersect at right angles at O. In triangles AOD and COD, AO=OC (diagonals bisect each other) OD=OD (common side) AD = DC i.e., the adjacent sides are equal. By definition, ABCD is a rhombus. Hence the theorem is prov..

i) AD is the median i.e., BD = DC In D ABK, F is the mid-point of AB . (given) G is the mid-point of AK. (by construction) FG || BK or GC || BK ...(1) By theorem on line joining the mid-points of any two sides of a triangle. Similarly in D ACK, GE || CK or GB || CK From (1) and (2)..