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As shown in the figure, point I is the heart of △ ABC, the extension line of AI intersects with BC at point D, and intersects with the circumscribed circle of △ ABC at point E. verify EB = EC = EI
EB = EC is obvious, equiangular to equilateral
Because I'm the heart
BAI=CAI
ABI=CBI
BIE=BAI+ABI=CAI+CBI=ABC
So the triangle BIE is isosceles triangle
EI=EB
EB=EC=EI
As shown in the figure, point I is the heart of △ ABC, the extension line of AI intersects at point D, and the circumcircle of △ ABC intersects at point E. (1) prove that IE = be; (2) if ie = 4, AE = 8, find the length of de
(1) It is proved that: connecting IB. ∵ point I is the heart of △ ABC, ∵ bad = ∵ CAD, ∵ ABI = ∵ IBD = ∵ ABI = ∵ CAD + ∵ IBD = ∵ IBD + ∵ DBE = ∵ IBE, ∵ be = ie. (2) in △ bed and △ AEB, ∵ EBD = ∵ CAD = ∵ bad, ∵ bed = ∵ AEB. ∵ bed ∽ AEB, ∵ beae = debe, ∵ ie = 4, AE = 8, ∵ be = 4, namely de = be2ae = 2
As shown in the figure, I is the heart of △ ABC, AI intersects BC with D, intersects △ ABC circumscribed with E, is judgment IE.BE The size of the relationship, and explain the reasons
-------------A-----
'
'
'--------I
'
B-----D------------C
'
'
'
--E
Equal
According to the title, ∠ BIE = ∠ Bei + ∠ EBI, and I is △ ABC, so ∠ Bei = ∠ EIC = ∠ EBC
Similarly, there is ∠ EBI = ∠ IBC, so ∠ EBI + ∠ Bei = ∠ EBC + ∠ IBC, i.e. ∠ BIE = ∠ EBI
So ie = be
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