Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 6.2 National olympiad Prove it Belarus

Given a trapezium $ABCD$ ( $AD \parallel BC$ ) with $AD = 3BC$ . A circle $\Gamma_1$ with the center $B$ passes through the midpoint of $BD$ , and a circle $\Gamma_2$ with the center $C$ passes through the midpoint of $AC$ .
Prove that the line through the points of intersection of $\Gamma_1$ and $\Gamma_2$ meets the side $AD$ at its midpoint.

Solution

Let $R$ be the midpoint of $BD$ , $M$ be the midpoint of $AD$ , $P$ be the point of intersection of $AC$ and $BD$ . We show that $LD = 2LP$ .

By condition, $BP : PD = 1 : 3$ . So $BP = PR = 0.5RD$ , hence $LP$ is the median of the triangle $BLP$ , and $LR$ is the median of the triangle $BLD$ .

Using the formula for the median length, we obtain
$4LP^2 = 2LB^2 + 2LR^2 - BR^2, \quad 4LR^2 = 2LB^2 + 2LD^2 - BD^2.$
Thus
$\begin{aligned} 4LP^2 &= 2LB^2 + 2LR^2 - BR^2 = 2LB^2 + 0.5(2LB^2 + 2LD^2 - BD^2) - BR^2 = \\ &= 3LB^2 + LD^2 - 0.5(2BR)^2 - BR^2 = LD^2 - 3(BL^2 - BR^2) = [BL = BR] = LD^2. \end{aligned}$
Similarly, $LA = 2LP$ . Therefore, $LA = LD$ , i.e. $L$ lies on the perpendicular bisector of $AD$ . Similarly, $K$ lies on the perpendicular bisector of $AD$ . Hence the line $LK$ meets $AD$ at its midpoint.

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