Answer to Question #140802 in Geometry for Ali Atwi

Question #140802
ABC and VDC are two isosceles triangles drawn on both sides of BC .E is the symmetric of C with respect to A and F is the symmetric of C with respect to D.
Show that E,B and F are collinear
1
Expert's answer
2020-10-27T17:25:15-0400


Since EE is symmetric With respect to AA , AC=AEAC=AE , similarly to CB=ABCB=AB .

Let's lower the medians from the vertices AA and DD to the base BCBC , these will be the medians AHAH and DHDH , so ADAD is perpendicular to CBCB and divides it in half. ADAD is the middle line of ECF\triangle ECF (since AA and DD are the midpoints of ECEC and CFCF , respectively), hence CBCB is perpendicular to EFEF, since the ratio of the triangles CADCEF=12\frac{\triangle CAD}{\triangle CEF} = \frac{1}{2} , and CHCB=12\frac{CH}{CB}=\frac{1}{2} , then B belongs to the straight EFEF , hence B,E,FB, E, F are collinear.


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