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Let-M-be-a-point-in-interior-of-ABC-Three-lines-are-drawn-through-M-parallel-to-triangle-s-sides-thereby-producing-three-trapezoids-Suppose-a-diagonal-is-drawn-in-each-trapezoid-in-such-a-way-tha




Question Number 16951 by Tinkutara last updated on 28/Jun/17
Let M be a point in interior of ΔABC.  Three lines are drawn through M,  parallel to triangle′s sides, thereby  producing three trapezoids. Suppose a  diagonal is drawn in each trapezoid in  such a way that the diagonals have no  common endpoints. These three  diagonals divide ABC into seven  parts, four of them being triangles.  Prove that the area of one of the four  triangles equals the sum of the areas  of the other three.
$$\mathrm{Let}\:{M}\:\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{in}\:\mathrm{interior}\:\mathrm{of}\:\Delta{ABC}. \\ $$$$\mathrm{Three}\:\mathrm{lines}\:\mathrm{are}\:\mathrm{drawn}\:\mathrm{through}\:{M}, \\ $$$$\mathrm{parallel}\:\mathrm{to}\:\mathrm{triangle}'\mathrm{s}\:\mathrm{sides},\:\mathrm{thereby} \\ $$$$\mathrm{producing}\:\mathrm{three}\:\mathrm{trapezoids}.\:\mathrm{Suppose}\:\mathrm{a} \\ $$$$\mathrm{diagonal}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{in}\:\mathrm{each}\:\mathrm{trapezoid}\:\mathrm{in} \\ $$$$\mathrm{such}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{the}\:\mathrm{diagonals}\:\mathrm{have}\:\mathrm{no} \\ $$$$\mathrm{common}\:\mathrm{endpoints}.\:\mathrm{These}\:\mathrm{three} \\ $$$$\mathrm{diagonals}\:\mathrm{divide}\:{ABC}\:\mathrm{into}\:\mathrm{seven} \\ $$$$\mathrm{parts},\:\mathrm{four}\:\mathrm{of}\:\mathrm{them}\:\mathrm{being}\:\mathrm{triangles}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{four} \\ $$$$\mathrm{triangles}\:\mathrm{equals}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{areas} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{three}. \\ $$

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