As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 36 °, CD bisection ∠ ACB intersects AB at point D, AE parallels DC intersects BC extension line at point E, if DB = 2, CD = 3, AE = how much?

∵AB=AC,∠BAC=36°
∴∠ABC=∠ACB=72°
∵ CD bisection ∠ ACB
∴∠BCD=∠ACD=1/2∠ACB=36°
In △ ADC, it is isosceles triangle, ad = CD = 3
∵BD=2,AB=AD+DB=3+2=5
And ∵ AB = AC ∵ AC = 5
∵ CD parallel to AE
← EAC = ∠ ACD = 36 ° and ∠ AEC = ∠ BCD = 36 °
The Δ AEC is an isosceles triangle
∴EC=AC=5
∴BE=BC+EC=3+5=8
∵∠EAB=∠EAC+∠CAB=36°+36°=72°=∠ABC
The AEB is an isosceles triangle
∴AE=BE=8
That's right. How can it be different from the number of similar triangles?

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