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Math similar triangles in Grade Three.
Solution: ∫∠ADC =∠b+∠BAD =∠ADE+∠CDE

And < b = < ade = 45.

Bader =∠CDE

Similarly ∠B=∠C

∴ △BAD∽△CDE

∴ AB/DC=BD/CE

That is AB*CE=BD*DC.

∴ 2(2-Y)=X(2√2-X)

Therefore, the resolution function is: Y=X? /2-√2X+2(0 & lt; X & lt2√2)

When △ADE is an isosceles triangle, △ bad △ CDE

∴BD=CE, then X = 2-Y.

Substituting it into the above analytical formula for classification, we can get AE=Y=. .........