Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it United States

Problem:

Let ABCDEABCDE be a convex pentagon with CD=DECD = DE and ∠BCD=∠DEA=90∘\angle BCD = \angle DEA = 90^{\circ}. Point FF lies on ABAB such that AFAE=BFBC\frac{AF}{AE} = \frac{BF}{BC}. Prove that ∠FCE=∠ADE\angle FCE = \angle ADE and ∠FEC=∠BDC\angle FEC = \angle BDC.

Solution

Solution:

Let ω\omega denote the circumcircle of △CDE\triangle CDE and let D1D_{1} be the point opposite to DD. Let DADA meet ω\omega at A1A_{1} and let F′=A1C∩ABF' = A_{1}C \cap AB. If we let α=∠DA1C=∠DD1C=∠ED1D\alpha = \angle DA_{1}C = \angle DD_{1}C = \angle ED_{1}D then
AF′AE=AF′⋅AD1AA1⋅AD=sin⁡α⋅sin⁡∠D1DAsin⁡∠A1F′A⋅sin⁡α=sin⁡∠D1CAsin⁡∠CF′B=F′BBC \frac{AF'}{AE} = \frac{AF' \cdot AD_{1}}{AA_{1} \cdot AD} = \frac{\sin \alpha \cdot \sin \angle D_{1}DA}{\sin \angle A_{1}F'A \cdot \sin \alpha} = \frac{\sin \angle D_{1}CA}{\sin \angle CF'B} = \frac{F'B}{BC}
Hence F=F′F = F'. Now ∠F′CE=∠A1CE=∠A1DE=∠ADE\angle F'CE = \angle A_{1}CE = \angle A_{1}DE = \angle ADE.

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