Скачать или смотреть प्रमेय 6.1 वर्ग दशम, आधारभूत प्रमेय Thales' Theorem (The Basic Proportionality Theorem - BPT

​If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
​Illustration:
​Consider a \triangle ABC.
​Let a line DE be drawn such that DE is parallel to side BC (DE \parallel BC).
​The line DE intersects side AB at D and side E
The proof relies on the concept of area of triangles and similar triangles.
​Construct lines BE and CD.
​The triangles \triangle ADE and \triangle DBE share the same height (from E to AB), so their area ratio is equal to the ratio of their bases: \frac{Area(\triangle ADE)}{Area(\triangle DBE)} = \frac{AD}{DB}.
​Similarly, \frac{Area(\triangle ADE)}{Area(\triangle DEC)} = \frac{AE}{EC}.
​Since the base DE is common to \triangle DBE and \triangle DEC, and DE is parallel to BC, the tw
o triangles are between the same parallel lines, meaning they have the same height. Therefore, Area(\triangle DBE) = Area(\triangle DEC).
​By substituting the equal areas, we get the desired proportionality: \frac{AD}{DB} = \frac{AE}{EC}.



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