As shown in the figure, in △ ABC, CE and BF are two heights. If a = 70 ° and BCE = 30 °, then the degree of EBF is______ The degree of FBC is______ ．

As shown in the figure, in △ ABC, CE and BF are two heights. If a = 70 ° and BCE = 30 °, then the degree of EBF is______ The degree of FBC is______ ．

In RT △ ABF, ∠ a = 70, CE, BF are two high, ∫ EBF = 20 °, ∫ ECA = 20 ° and ∫ BCE = 30 °, ∫ ACB = 50 °, ∫ FBC = 40 ° in RT △ BCF

In the triangle ABC, angle a = 60 degrees, e and F are the points on AB and AC respectively, and angle ECB = angle FBC = 30 degrees. The intersection of BF and CE is d. try to explain that D is three
The center of the circumscribed circle.

Because angle ECB = angle FBC = 30 degrees,
So BDC = 180 degrees - ECB - FBC = 120 degrees,
Because angle a = 60 degrees, that is angle BDC = 2 angle a,
So D is the center of circumcircle of angle triangle ABC

As shown in the figure, in the triangle ABC, AB > AC, make the angle FBC = angle ECB = 1 / 2 angle a, CE and BF intersect at point m, and verify: be = BF

Extend me to p to make MP = MF
∵∠FBC=∠ECB=x
∴△PMB≌△FMC
∵∠FMC=2x=∠BAC
∴∠AEC=∠MFC=∠PEB=∠MPB
∴PB=EB
∵PB=FC
∴BE=CF
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