It is known that in the triangle ABC, AB is greater than AC, the bisector of angle B intersects d with the bisector of the outer angle of angle c, DF parallels BC, and AB and AC intersect F and e respectively BF=EF+CE.
Set point G on the extension line of BC
Known, DF ‖ BC,
It can be concluded that: ∠ FDB = ∠ CBD = ∠ FBD, ∠ EDC = ∠ GCD = ∠ ECD,
So BF = DF, CE = De,
BF = DF = EF + de = EF + CE
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- 1. As shown in Figure 5, it is known that in the triangle ABC, AB is greater than AC, the bisector of angle B and the bisector of the outer angle of angle c intersect at D and DF, and BC intersects AB and AC at f and e respectively Question: explain BF = EF + CE.
- 2. As shown in the figure, it is known that de ∥ BC, DF and be are divided into ∥ ade and ∥ ABC equally, and it is proved that ∥ FDE = ∥ DEB
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