Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it New Zealand

Problem:

Let ABCABC be an isosceles triangle with AB=ACAB = AC. Point DD lies on side ACAC such that BDBD is the angle bisector of ABC\angle ABC. Point EE lies on side BCBC between BB and CC such that BE=CDBE = CD. Prove that DEDE is parallel to ABAB.

Solution

Solution:

Let EE' be the point on line BCBC such that DEDE' is parallel to ABAB. We know that EE' lies between BB and CC because DD lies between AA and CC. So it suffices for us to prove that BE=CDBE' = CD.

Figure 1

ACB=CBA=CED\angle ACB = \angle CBA = \angle CE'D.

Therefore triangle CDECDE' is isosceles with CD=DECD = DE'.

EBD=DBA=ADE\angle E'BD = \angle DBA = \angle ADE'.

Therefore triangle BEDBE'D is isosceles with BE=DEBE' = DE'. Hence CD=DE=BECD = DE' = BE' as required.

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