Proving Theorems
Proving theorems are obtained from deductive reasoning, rather than from inductive or observed point of view. Proof should make obvious that a declaration is true in all cases, without a particular exclusion. An original suggestion that is supposed to be true is known as a conjecture. The declaration that is proved is often called a theorem. A theorem may be referred to as a lemma.
Theorem 1: Angle in the segment of a similar circle is equivalent.
Proof:
Segment of circle:
The section in the internal of a circle is separated into two parts by the chord AB called segments. The segment that contains the centre is the major segment and the smaller section is the minor segment. A diameter divides the boundary into two equal parts. Each of these is called a semicircle.
Let us now look at the case once an arc AB is a semicircle. We join CO and produce it to P. We note that,
<AOP + <BOP = 180 degree
<AOB = 180 degree
But <AOB = 2 <ACB
<ACB = ½ <AOB
= ½ * 180 degree = (180 / 2 ) degree
= 90 degree= a right angle
Hence the theorem is proving.
Theorem 2: Proving the theorem correct angle from the centre of a circle to a triad bisect the triad.
Given:
AB is a triad in a circle among centre O. OC ┴ AB.
To proving:
The point C bisects the chord AB.
Construction:
Join OA and OB
Proof:
In triangles OAC and OBC
<OCA = <OCB = 90degree ; OA = OB (Radii)
OC = OC
ΔOAC ≡ ΔOBC (RHS)
CA ≡ CB
The point C bisects the chord AB.
Hence the theorem is proving by CA=CB.
Basic proportionality theorem: line is drawn equivalent to single side of a triangle then the other two sides are divided in the equal ratio.
Proof:
Draw a line AP. Beginning the point A let us step off five equivalent segments on the line AP. Let us denote 2 and 5 points by B and C correspondingly. In the course of A let us draw a line AD. In the course of B and C let as draw lines BE, CF both parallel to the line AD.
In ΔACH , BG||CH.
Then AB / BC = AG /GH
Adding 1 on both sides,
(AB /BC)+1 = (AG/GH)+1
Or
(AB+BC)/BC= (AG+GH)/GH
Or
AC /BC=AH/GH
Similarly we can get the result
AB /AC = AG/ AH
Hence the theorem is proving.