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Prove that the line which divides the nonparallel sides of a trapezium proportionally is parallel to the third side.?
Wiki User
∙ 11y ago
Theorem 3 : Any line parallel to the sides of a trapezium (trapezoid) divides the non-parallel sides proportionally.
Given : ABCD is a trapezoid. DC AB. EF AB and EF DC.
Prove that : AE/ED = BF/FC
Construction : Join AC, meeting EF in G.
StatementsReasons1) EG DC1) Given (in ΔADC)2) AE/ED = AG/GC2) By Basic proportionality theorem3) GF AB3) Given (in ΔABC)4) AG/GC = BF/FC4)By Basic proportionality theorem5) AE/ED = BF/FC5) From (2) and (4)
Source: ask-math.com
Wiki User
∙ 11y ago
To prove that the line which divides the nonparallel sides of a trapezium proportionally is parallel to the third side, we can use the property of similar triangles. Let the trapezium ABCD have sides AB and CD as the nonparallel sides, and side BC as the third side. Let the line dividing AB and CD be denoted as EF, with E on AB and F on CD. By the property of similar triangles, we can show that triangles AEF and BCF are similar, and hence their corresponding angles are congruent. This proves that EF is parallel to BC.
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