Solutions - Georg Mohr-Konkurrencen

̸AB ‖ EC and ED ‖ AC , we have ̸ CAB = ̸ DEC and the arcs DCand BC are of equal length. Since P E is tangent to c and |AE| = |DC|,then ̸ P EA = ̸ DBC = ̸ QBC . As ABCD is inscribed in c, we have̸ QCB = 180 ◦ − ̸ EAB = ̸ P AE . Also, ABCD is an isosceles trapezium,whence |AE| = |BC|. So the triangles AP E and CQB are congruent, and|QC| = |P A|. Now P ACQ is a quadrilateral with a pair of opposite sidesequal and parallel. So P ACQ is a parallelogram, and |P Q| = |AC|.7. Let X be the point on segment AC such that ̸ ADX = ̸ AKN , then̸ AXD = ̸ ANK = 180 ◦ − ̸ AMK(see Figure 2). Triangles NAK and XAD are similar, having two pairs of|AN| · |AD|PSfrag replacements equal angles, hence |AX| = . Since triangles MAK and XCD|AK|A|AM| · |CD| |AM| · |AB|are also Bsimilar, we have |CX| = = and|AK| |AK|C|AM| D· |AB|+|AN| · |AD|=(|AX|+|CX|) · |AK|=|AC| · |AK| .EQDCPNKXAMBFigure 28. Let X be the point symmetric to B with respect to AN , and let Y be thepoint symmetric to C with respect to DN (see Figure 3). ThenXNY = 180 ◦ − 2 · (180 ◦ − 135 ◦ ) = 90 ◦and |NX| = |NY | = |BC| . Therefore, |XY | = |BC| √ . Moreover, we have226