, 2010

Let $a = |AB|$ and $b = |CD|$ and let $h_a$ and $h_b$ be the altitudes of triangles $PAB$ and $PCD$ drawn from $P$ (see Fig. 7). Denote $S_1 = S_{PAB} + S_{PCD}$ and $S_2 = S_{PBC} + S_{PDA}$. Then $S_1 = \frac{1}{2}(a h_a + b h_b)$ and $S_1 + S_2 = \frac{1}{2}(a + b)(h_a + h_b)$, whence $S_2 = \frac{1}{2}(a h_b + b h_a)$. Since triangles $PAB$ and $PCD$ are similar, $a > b$ implies $h_a > h_b$ and also $a < b$ implies $h_a < h_b$ ($a \neq b$ because $a$ and $b$ are the lengths of the bases of the trapezoid). Hence
$$S_1 - S_2 = \frac{1}{2}(a h_a + b h_b - a h_b - b h_a) = \frac{1}{2}(a - b)(h_a - h_b) > 0,$$
i.e., $S_1 > S_2$.

Let $M$ and $N$ be the intersection points of the arms $BC$ and $DA$ of the trapezoid with the line being parallel to the bases of the trapezoid and passing through point $P$. Let $l$ be the length of $MN$, let $d$ be the length of the midline of the trapezoid, and let $h$ and $S$ be the height and the area of the trapezoid, respectively. Let $S' = S_{PBC} + S_{PDA}$. Then $S = d h$ and $S' = \frac{1}{2} l h$ whence it suffices to show that $l < d$.

Fig. 7

W.l.o.g., assume $|AB| < |CD|$. Comparing the heights of similar triangles $PAB$ and $PCD$ shows that $MN$ is closer to base $AB$ than to base $CD$. Thus $MN$ is situated between the midline and the shorter base $AB$. Consequently, $MN$ is shorter than the midline.

Let $a = |AB|$ and $b = |CD|$. Let $h$ and $S$ be the height and the area of the trapezoid, respectively, and let $h_a$ and $h_b$ be the heights corresponding to vertex $P$ of triangles $PAB$ and $PCD$, respectively. Similar triangles $PAB$ and $PCD$ imply $h_a : h_b = a : b$. As $h_a + h_b = h$ we get
$$h_a = \frac{a}{a + b} \cdot h, \quad h_b = \frac{b}{a + b} \cdot h.$$
Now
$$S_{PAB} + S_{PCD} = \frac{1}{2}(a h_a + b h_b) = \frac{1}{2} \cdot \frac{a^2 + b^2}{a + b} \cdot h.$$
It suffices to show that $S_{PAB} + S_{PCD} > \frac{S}{2}$ or, equivalently,
$$\frac{1}{2} \cdot \frac{a^2 + b^2}{a + b} \cdot h > \frac{1}{2} \cdot \frac{a + b}{2} \cdot h,$$
or, equivalently, $2(a^2 + b^2) > (a + b)^2$. But the last inequality is equivalent to $(a - b)^2 > 0$ ($a \neq b$ since $a$ and $b$ are the lengths of the bases of the trapezoid).