In the triangle ABC, point D is a point on the ray Ba, make de = CD, intersect the straight line BC at point E. when point D is on AB, find CE = AD + AC These triangles are equilateral
prove
Make DF ‖ BC through point D and pass CA to point F
Then, FDC = DCE, f = ACB = 60 degree, ADF = b = 60 degree
A triangle is an equilateral triangle
∴∠CFD=∠DBE=120°,DF=AD
∵DE=DC
∴∠E=∠DCE
∴∠E=∠FDC
∴△FDC≌△BED
∴BE=FD
∴BE=AD
∴CE=BE+BC=AD+AC
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