Before moving to finding the middle line of the triangle, you need to recall the second sign of the similarity of the triangles and the properties of the direct parallelism.
How to find the middle line of the triangle - the second sign of the similarity of triangles
Figure 1 shows two triangles. ABC triangle is similar to the A1B1C1 triangle. And the adjacent parties are proportional to, that is, AB refers to A1B1 as well as AC refers to A1C1. These two conditions and follow the similarity of triangles.
How to find the middle line of the triangle - a sign of parallelism of direct
Figure 2 shows direct a and b, sequential c. At the same time, 8 corners are formed. Corners 1 and 5 respective, if straight parallel, then corresponding angles are equal, and vice versa.
How to find the middle line of the triangle
In Figure 3, M of the middle of AB, and n middle AC, BC base. Cut MN - called the middle line of the triangle. The same theorem says - the middle line of the triangle is parallel to the base and is equal to its half.
In order to prove that Mn is the middle line of the triangle, we will need the second sign of the similarity of triangles and a sign of direct parallelism.
The AMN triangle is similar to the ABC triangle, on the second basis. In such triangles, corresponding angles are equal, angle 1 is equal to the angle 2, and these angles are appropriate with the intersection of two direct secant, therefore, directly parallel, Mn in parallel BC. Angle A Common, AM / AB \u003d AN / AC \u003d ½
The likeness ratio of these triangles ½, it follows from this that ½ \u003d Mn / Bc, Mn \u003d ½ BC
So we found the middle line of the triangle, and proved the theorem about the middle line of the triangle, if you still do not understand how to find the average line, watch the video below.
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