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 E′E' be the point on line BCBC such that DE′DE' is parallel to ABAB. We know that E′E' 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=∠CE′D\angle ACB = \angle CBA = \angle CE'D.

Therefore triangle CDE′CDE' is isosceles with CD=DE′CD = DE'.

∠E′BD=∠DBA=∠ADE′\angle E'BD = \angle DBA = \angle ADE'.

Therefore triangle BE′DBE'D is isosceles with BE′=DE′BE' = DE'. Hence CD=DE′=BE′CD = 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.