
1. Let $A B C D$ be a convex quadrilateral with $A B=a, B C=b, C D=c$ and $D A=d$. Suppose

$$
a^{2}+b^{2}+c^{2}+d^{2}=a b+b c+c d+d a
$$

and the area of $A B C D$ is 60 square units. If the length of one of the diagonals is 30 units, determine the length of the other diagonal.


## Solution

$a^{2}+b^{2}+c^{2}+d^{2}=a b+b c+c d+d a \Rightarrow(a-b)^{2}+(b-c)^{2}+(c-d)^{2}+(d-a)^{2}=0 \Rightarrow a=b=c=d$. Thus $A B C D$ is a rhombus and

$$
[A B C D]=(1 / 2)\left(d_{1} d_{2}\right)
$$

where $d_{1}$ and $d_{2}$ are the lengths of the diagonals. Hence $d_{2}=\frac{2[A B C D]}{d_{1}}=4$ units.
