Let's introduce notations: Problem. Let ABCD be a convex quadrilateral with the area 32, and assume that AB + BD + CD = 16. Find AC. Solution. Let S and S' be the areas of triangles ABD and CDB, respectively. Then, the area of the quadrilateral ABCD is obviously the sum of these two areas S and S'; since we know that the area of the quadrilateral ABCD is 32, we thus get 32 = S + S'. Since the area of a triangle equals half the product of two sides multiplied with the sine of the angle between them, the area S of triangle ABD must be $S=\frac12\cdot AB\cdot BD\cdot\sin\measuredangle ABD$. Similarly, $S^{\prime}=\frac12\cdot CD\cdot BD\cdot\sin\measuredangle CDB$. Hence, $32=S+S^{\prime}=\frac12\cdot AB\cdot BD\cdot\sin\measuredangle ABD+\frac12\cdot CD\cdot BD\cdot\sin\measuredangle CDB$. Now, we have $\sin\measuredangle ABD\leq 1$ with equality if and only if < ABD = 90°, and we have $\sin\measuredangle CDB\leq 1$ with equality if and only if < CDB = 90°. Also, by the AM-GM inequality, applied to the numbers AB + CD and BD, we get $\left(AB+CD\right)\cdot BD\leq\left(\frac{\left(AB+CD\right)+BD}{2}\right)^2=\left(\frac{AB+BD+CD}{2}\right)^2$ $=\left(\frac{16}{2}\right)^2=64$. Hence, $32=\frac12\cdot AB\cdot BD\cdot\sin\measuredangle ABD+\frac12\cdot CD\cdot BD\cdot\sin\measuredangle CDB$ $\leq\frac12\cdot AB\cdot BD\cdot 1+\frac12\cdot CD\cdot BD\cdot 1=\frac12\cdot\left(AB+CD\right)\cdot BD\leq\frac12\cdot 64=32$. Thus, we have a chain of inequalities with the very left and the very right hand side being equal. This means that all the inequalities inbetween must be equalities. Now, the inequality $\sin\measuredangle ABD\leq 1$ becomes an equality if and only if < ABD = 90°, the inequality $\sin\measuredangle CDB\leq 1$ becomes an equality if and only if < CDB = 90°, and the inequality $\left(AB+CD\right)\cdot BD\leq 64$, being derived from the AM-GM inequality applied to the numbers AB + CD and BD, becomes an equality if and only if AB + CD = BD. Hence, we must have < ABD = 90°, < CDB = 90° and AB + CD = BD. Now, < ABD = 90° yields $AB\perp BD$, and < CDB = 90° yields $CD\perp BD$; hence, AB || CD. Thus, by Thales, $\frac{AB}{CD}=\frac{BR}{DR}$, where R is the point of intersection of the segments AC and BD. Hence, $\frac{BD}{CD}=\frac{AB+CD}{CD}=\frac{AB}{CD}+1=\frac{BR}{DR}+1=\frac{BR+DR}{DR}=\frac{BD}{DR}$. Hence, CD = DR, so the triangle CDR is an isosceles right triangle (right since < CDR = < CDB = 90°). Hence, its hypotenuse CR equals $CR=\sqrt2\cdot DR$. Similarly, $AR=\sqrt2\cdot BR$. Now, since AB + CD = BD, we have $16=AB+BD+CD=\left(AB+CD\right)+BD=BD+BD=2\cdot BD$, so that $BD=\frac{16}{2}=8$. Thus, $AC=AR+CR=\sqrt2\cdot BR+\sqrt2\cdot DR=\sqrt2\cdot\left(BR+DR\right)=\sqrt2\cdot BD=\sqrt2\cdot 8$. Darij

I found a very short solution, Notation: Let $ABCD$ be the quadrilateral with the area 32 with AB + BD + CD = 16, let $h_a$ and $h_c$ the length of the altitudes from $A,C$ to $BD$. $$ h_a\cdot BD+ h_c\cdot BD=64 $$$$ BD^2+h_a\cdot BD+ h_c\cdot BD=BD^2+64 $$$$ BD(BD+h_a+h_c)=BD^2+64 $$ Now, $AB\geq h_a$ and $CD\geq h_c$, then $16\cdot BD\geq BD^2+64$, this implies $BD=8$ and $h_a=AB, h_c=CD$, the rest is easy, $AC$ is a diagonal of a 8-side square then $AC=\sqrt{2}\cdot 8$.