# A Property of Trapezoid with a Right Angle

### Problem

In trapezoid $ABCD,\;$ with $AB\parallel CD\;$ and $AB\perp AD,\;$ $DN\perp AC\;$ and $AM\perp BD;\;$ $E\;$ the intersection of $DN\;$ and $AM;\;$ $EF\perp AD;\;$ $O\;$ the midpoint of $AD,\;$ $P\;$ that of $EF.$

Prove that $OP\perp BC.$

### Solution 1

WLOG, assume $A=(-u,0),\;$ $D=(v,0),\;$ $E=(0,1).\;$ Clearly, $O=\displaystyle\left(\frac{v-u}{2},0\right)\;$ and $P=\displaystyle\left(0,\frac{1}{2}\right),\;$ because $EF\;$ is the third altitude in $\Delta ADE.$

We also get he following equations:

$\begin{cases} AE:& -x+uy=u,\;\text{implying}\\ DM:& ux+y=ub\;\text{and, since}\\ AB:& x=-u,\;\text{then}\;B=(-u,u(u+v));\\ AD:& x+vy=v,\;\text{implying}\\ EN:& -vx+y=uv,\;\text{and, since}\\ CD:& x=v,\;\text{then}\;C=(v,v(u+v)). \end{cases}$

It follows that $\overrightarrow{BC}=(u+v)[\overrightarrow{i}+(v-u)\overrightarrow{j}].\;$ But $\displaystyle\overrightarrow{OP}=\frac{1}{2}[(v-u)\overrightarrow{i}-\overrightarrow{j}].\;$ From here, $\overrightarrow{BC}\cdot\overrightarrow{OP}=0,\;$ i.e., $OP\perp BC.$

### Solution 2

We have to show that $\displaystyle\frac{AB-CD}{AD}\;$ and $\displaystyle 2\frac{OF}{EF}=\frac{AF-FD}{EF}\;$ are equal.

But it follows immediately from $\displaystyle\frac{AF}{EF}=\frac{AB}{AD}\;$ and $\displaystyle\frac{FD}{EF}=\frac{CD}{AD}\;$ (equal angles between pairwise perpendicular lines) that $\displaystyle\frac{AF-FD}{EF}=\frac{AB-CD}{AD}.$

### Acknowledgment

The problem, due to Elberling Vargas Diaz, has been kindly communicated to me by Leo Giugiuc, along with his solution above. Solution 2 is by Grégoire Nicollier.