geometry
This problem is hard for me. Can anyone help me? Thank you in advance.
Let ABCD be a trapezoid with the measure of base AB twice that of base DC, and let E be the point of intersection of diagonals. If the measure of diagonal AC is 11, then that of segment EC is equal to?
Ok, here is one approach that works. Since diagonals are equal, and bases are parallel, it is easy to show that top and bottom triangle are similar, using alternate interior and vertical angles theorem. Let's call the diagonal intersection point S and vertices of the trapezoid A,B,C,D with A at bottom left vertex. Now BS is a multiple of DS (say factor k) and AS is a multiple of CS (same factor k) Now use area formula for triangle ½*b*c*sin(enclosed angle) on left and right triangle to show that these areas are equal. You will in both cases arrive at area ½*AS*DS*sin(angle) for left and right triangle of the trapezoid. KNowing that these triangles are congruent, and top and bottom triangles are similar, I think you can finish the proof
Assume that the diagonals cut themselves in four segments with lengths $a,b,c,d$, in such a way that $b+d=10,a+c=8$ and the right triangle with legs $a,b$ is similar to the right triangle with legs $c,d$, so that $\frac{b}{a}=\frac{d}{c}$ or, equivalently, $ad=bc$. That leads to the equation: $$ a(10-b)=b(8-a) $$ that is equivalent to $10a=8b$, hence $\frac{b}{a}=\frac{d}{c}=\frac{5}{4}$. So we have that all the trapezoids with perpendicular diagonals such that $a=x,\,b=\frac{5}{4}x,\,c=y,\,d=\frac{5}{4}y,\,x+y=8$ meet the given constraints. For such trapezoids, the bases have length $$ \frac{x}{4}\sqrt{41},\qquad \frac{y}{4}\sqrt{41} $$ by the Pythagorean theorem, hence the "median" has length $$ \frac{x+y}{8}\sqrt{41} = \color{red}{\sqrt{41}}.$$
In general, if a trapezoid has perpendicular diagonals with length $\ell_1,\ell_2$, the segment joining the midpoints of the diagonal sides has length $\frac{1}{4}\sqrt{\ell_1^2+\ell_2}$. Here it comes a proof (almost) without words: the red segments have the same length.
Best Answer
triangles ABE and CDE are similar, hence: $\frac{AB}{DC}=\frac{AE}{CE}=2$ and AE+EC=11. Hence, EC=$11/3$.