The figure shows a parallelogram ABCD so that BF = DE. Lines BE and DF meet at G. Prove that CG bisects the angle BCD. This entry contributed by Ajit Athle.
Here, a really cool parallelogram phenomenon not often seen. How to prove? 🤔Source: @gogeometry. https://t.co/znQa16wtXt @geogebra #MTBoS #ITeachMath #geometry #proof #math #maths #EdTech #HSMath #CollegeMath #MathEd pic.twitter.com/frXlbX2Lcr— Tim Brzezinski (@Brzezinski_Math) December 12, 2019
Here, a really cool parallelogram phenomenon not often seen. How to prove? 🤔Source: @gogeometry. https://t.co/znQa16wtXt @geogebra #MTBoS #ITeachMath #geometry #proof #math #maths #EdTech #HSMath #CollegeMath #MathEd pic.twitter.com/frXlbX2Lcr
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