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"Evidence by Contradiction" PPT Courseware 3
A circle cannot be drawn through three points on the same straight line.
It is known that: points A, B, and C are on the straight line L.
Prove: A circle cannot be drawn through three points A, B, and C.
Proof: Suppose a circle can be drawn through three points A, B, and C.
Suppose the center of this circle is P, then point P is both on the perpendicular bisector L1 of line segment AB and on the perpendicular bisector L2 of line segment BC, that is, point P is the intersection of L1 and L2.
And this contradicts what we have learned before: "There is and is only one straight line perpendicular to the known straight line through a point." Assumption is not true.
Therefore, a circle cannot be drawn through three points on the same straight line.
This method of proving a proposition is called proof by contradiction.
The general steps to prove that a proposition is true by contradiction are:
The first step is to assume that the proposition is not true.
The second step is to start from this assumption and other known conditions, and through reasoning and argumentation, arrive at results that are contradictory to the learned concepts, basic facts, proven theorems, properties, or problem conditions.
The third step is to determine from the contradictory results that the hypothesis is not established, thereby indicating that the conclusion of the proposition is correct.
Cooperative learning
Prove: In the same plane, if two straight lines are parallel to a third straight line, then the two straight lines are also parallel to each other.
(1)Which proof method will you choose first?
(2) If you choose proof by contradiction, what should you assume first? What does the result conflict with?
Known: As shown in the figure, l1∥l2 ,l 2 ∥l 3
Prove: l1∥l３
Proof: Suppose l1 is not parallel to l3, then l1 and l3 intersect, and let the intersection point be p.
∵l1∥l2, l2∥l3, then there are two straight lines l1 and l3 passing through the point p, both of which are parallel to l2. This is contradictory to "passing a point outside the straight line, there is and is only one straight line parallel to the known straight line".
Therefore, the hypothesis does not hold, and the conclusion verified holds,
That is l1∥l３
give it a try
Known: As shown in the figure, straight lines a and b are intercepted by straight line c, ∠1 ≠ ∠2. Prove: a∥b
Proof: Assuming that the conclusion does not hold, then a∥b
∴∠1=∠2 (two straight lines are parallel and have equal angles)
This contradicts the known ∠1≠∠2
∴Assumption is not true
∴a∥b
Extend and expand
Can you prove the following proposition by proof by contradiction?
As shown in the figure, in △ABC, if ∠C is a right angle, then ∠B must be an acute angle.
Proof: Assuming that the conclusion is not true, then ∠B is a right angle or an obtuse angle.
When ∠B is a right angle, then ∠B+ ∠C= 180°
This contradicts the fact that the sum of the three interior angles of a triangle is equal to 180°;
When ∠B is an obtuse angle, then ∠B+ ∠C＞180°
This contradicts the fact that the sum of the three interior angles of a triangle is equal to 180°;
In summary, the hypothesis is not valid.
∴∠B must be an acute angle.
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