# Signs of parallel lines

The parallelism of two straight lines can be proved on the basis of the theorem, according to which, two perpendiculars drawn with respect to one straight line, will be parallel. There are certain signs of parallelism of straight lines - there are three of them all, and we will consider all of them more specifically.

## The first sign of parallelism

Straight lines are parallel if, at the intersection of their third straight line, the inner corners formed, which are crossed, are equal.

Suppose, at the intersection of lines AB and CD with a straight line �F, angles / 1 and / 2 were formed. They are equal, since the straight line EF runs under one bias with respect to the other two straight lines. At the intersection of the lines, we put the points of Ki L - we have a segment of the secant EF. Find its middle and put the point O (Fig. 189).

We lower the perpendicular from the point O to the line AB. Let's call it OM. Continue perpendicular until it intersects with direct �D. As a result, the original line AB is strictly perpendicular to MN, which means that CD_ | _MN, but this statement requires proof.As a result of the perpendicular and the intersection line, we have formed two triangles. One of them is MY, the second is NOC. Consider them in more detail. signs of parallelism direct class 7

These triangles are equal, because, in accordance with the conditions of the theorem, / 1 = / 2, and in accordance with the construction of triangles, the side �K = to the side �L. MOL = / N�� angle, since these are vertical angles. From this it follows that the side and two angles adjacent to it of one of the triangles are respectively equal to the side and two angles adjacent to it, the other of the triangles. Thus, the MOL = triangle NOK and, therefore, the angle LMO = angle KNO, but we know that / LMO is a straight line, which means the corresponding angle, the angle KNO is also a straight line. That is, we managed to prove that to a straight line MN, both a straight line AB and a straight line CD are perpendicular. That is, AB and CD with respect to each other are parallel. This is what we needed to prove. Consider the remaining signs of parallelism of straight lines (class 7), which differ from the first characteristic in the method of proof.

## The second sign of parallelism

According to the second feature of parallelism of straight lines, we need to prove that the angles obtained in the process of intersection of parallel straight lines AB and CD of straight line EF will be equal.Thus, the signs of parallelism of two straight lines, both the first and second, are based on the equality of the angles obtained when they cross the third line. We assume that / 3 = / 2, and the angle 1 = / 3, since it is vertical to it. Thus, / 2 will be equal to angle 1, however, it should be taken into account that both angle 1 and angle 2 are internal, crosswise lying angles. Consequently, it remains for us to apply our knowledge, namely that the two segments will be parallel, if at their intersection of the third straight line the formed, crosswise lying angles will be equal. Thus, we found that AB || CD.

We managed to prove that under the condition of parallelism of two perpendiculars to one straight line, according to the corresponding theorem, the sign of parallelism of straight lines is obvious.

## The third sign of parallelism

There is also a third sign of parallelism, which is proved by the sum of one-sided interior angles. This proof of the parallel feature of straight lines allows us to conclude that the two straight lines will be parallel, if at the intersection of their third straight line, the sum of the obtained one-sided internal angles will be equal to 2d. See Figure 192.

### Related news

How to get political asylum

Deceased Russian actors, the good memory of which lives in our hearts

What is soy milk and what is its benefit

How many calories in carbohydrates

Promotion

The amount of the bank they want to load money the card will send me to ask for money

Work in a taxi

How to make a notebook