geometry
In isosceles trapezoid $ABCD$ the median $MN$ cuts diagonals $AC$ and $BD$ at $P$ and $Q$, respectively. If $PQ$ is $4$ find the perimeter of the trapezoid. I found that $AC=BD=8$ and I know that $PQ$ is half of $AB-CD$ but I can't find $CD$. Any help would be appreciated.
Angle A is 60 degrees. $MN$ is 8.
Edit: I wasn't given that $MN$ is 8. Now when I know that I can find the perimeter. $P=a+b+2(a-b)=32$
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
The problem is ill-posed; the perimeter isn't determined by the given information. If we move $D$ and $C$ outward, as they come to lie vertically above $A$ and $B$, respectively, $P$ and $Q$ converge at the centre. Thus the trapezoid has to be scaled up without bounds for $PQ$ to retain the value $4$, and thus it can have arbitrarily large perimeters.