How to find the middle line of a trapezoid?
A segment of a straight line connecting the midpoints of the sides of a trapezium is called the middle line of the trapezium. How to find the center line of a trapezoid and how it relates to other elements of this figure, we will describe below.
Draw a trapezoid in which AD is a larger base, BC is a smaller base, EF is the middle line. Extend the base AD beyond the point D. Let us draw the line BF and continue it to the intersection with the extension of the base AD at the point O. Consider the triangles ∆BCF and ∆DFO. Angles ∟BCF = ∟DFO as vertical. CF = DF, ∟BCF = ∟FDO, since Sun // JSC. Therefore, the triangles ∆BCF = ∆DFO. Hence the sides bf = fo.
Now consider ∆ABO and ∆EBF. ∟ABO is common to both triangles. BE / AB = ½ by condition, BF / BO = ½, since ∆BCF = ∆DFO. Therefore, the triangles ABO and EFB are similar. Hence the ratio of the sides EF / AO = ½, as well as the ratio of other parties.
Find EF = ½ AO. The drawing shows that AO = AD + DO. DO = BC as sides of equal triangles, so AO = AD + BC. Hence, EF = ½ AO = ½ (AD + BC). Those. the length of the trapezium midline is equal to the half-sum of the bases.
Is the center line of a trapezoid always equal to half the sum of the bases?
Suppose that there is a special case where EF ≠ ½ (AD + BC). Then BC ≠ DO, therefore, ∆BCF ≠ ∆DCF. But this is impossible, since they have two angles and the sides between them. Therefore, the theorem is true under all conditions.
Medium Line Problem
Suppose in our trapezoid ABCD AD // BC, ∟A = 90 °, ∟C = 135 °, AB = 2 cm, the diagonal of the speaker is perpendicular to the side. Find the center line of the trapezoid EF.
If ∟A = 90 °, then ∟В = 90 °, then ∆ABC is rectangular.
∟BCA = ∟BCD - ∟ACD. ∟ACD = 90 ° by the condition, therefore, ∟BCA = BCD - ACD = 135 ° - 90 ° = 45 °.
If in a right-angled triangle ∆ABC, one angle is equal to 45 °, then the legs in it are equal: AB = BC = 2 cm.
Hypotenuse AC = √ (AB² + BC²) = √8 cm.
Consider ∆ACD. ∟ACD = 90 ° by condition. ∟CAD = ∟BCA = 45 ° as the angles formed by the secant parallel bases of the trapezium. Therefore, the legs AC = CD = √8.
Hypotenuse AD = √ (AC² + CD²) = √ (8 + 8) = √16 = 4 cm.
The center line of the trapezoid is EF = ½ (AD + BC) = ½ (2 + 4) = 3 cm.
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