Maths Olympiad Prep

Library / /1 of 29

Geometry Difficulty 4.3 AIME Prove it Croatia

Let ABCDABCD be a cyclic quadrangle. The perpendicular bisector of BC‾\overline{BC} meets the side AB‾\overline{AB} in point EE. The circle through EE, CC and the midpoint FF of the side BC‾\overline{BC} meets the side CD‾\overline{CD} in GG. Prove that AD⊥FGAD \perp FG. (D. Monk)

Solution

Let the lines ADAD and FGFG intersect at point HH.

Figure 1

As quadrilaterals ABCDABCD and EFCGEFCG are cyclic, it follows that
∠HDG=180∘−∠ADC=∠ABC,∠HGD=∠FGC=∠FEC=∠FEB=90∘−∠ABC. \begin{align*} \angle HDG &= 180^\circ - \angle ADC = \angle ABC, \\ \angle HGD &= \angle FGC = \angle FEC = \angle FEB = 90^\circ - \angle ABC. \end{align*}
Therefore
∠DHG=180∘−∠HDG−∠HGD=90∘ \angle DHG = 180^\circ - \angle HDG - \angle HGD = 90^\circ
so AD⊥FGAD \perp FG.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.