In the figure, is a $$ riangle ABC $$ in which, $$ \angle ACB =90\degree $$ , $$ AC=3$$ , $$ BC = 4$$ , and point $$D$$ , $$ E$$ are two points on the side of $$AB$$ $$($$ not coincide with $$D$$ , $$E$$ $$) $$ , and $$ \angle DCE = \angle B .$$

Question

In the figure, is a $$	riangle ABC $$  in which, $$   \angle ACB =90\degree $$ , $$ AC=3$$ , $$ BC = 4$$ , and point $$D$$ , $$ E$$  are two points on the side of $$AB$$  $$($$ not coincide with $$D$$ , $$E$$   $$) $$ , and $$ \angle DCE = \angle B  .$$

Uniteasy Admin · Accepted Answer

In $$	riangle CED$$ and $$	riangle BCD  $$
$$\because   \angle DCE = \angle B  $$ and they share the same angle $$\angle BDC $$
$$	herefore   $$ The two triangles are similar, and their corresponding sides are proportional.
$$\dfrac{DC}{BD}  = \dfrac{ED}{DC}  $$
$$DC^2 = BD\cdotp ED $$n

$$CD = \dfrac{AC\cdotp BC }{AB}  = \dfrac{12}{5}  $$
$$BD = \sqrt{BC^2-CD^2}  = \sqrt{16-(\dfrac{12}{5})^2  }  =\dfrac{16}{5}    $$
From (1),
$$ED =  \dfrac{DC^2}{BD}  = \dfrac{(\dfrac{12}{5})^2  }{\dfrac{16}{5}  } =\dfrac{9}{5}   $$
$$BE = BD-ED$$
$$=\dfrac{16}{5} - \dfrac{9}{5}  =\dfrac{7}{5}    $$ Since $$	riangle CED $$ 和 $$ 	riangle ACD  $$ are similar,  $$ 	riangle ACD  $$ is also an isosceles triangle
$$BD= BC = 4$$
Therefore, $$AD =AB-BD =1$$ $$AD = x$$ , $$ BE= y$$ 
Drop an altitude $$CF$$  from point $$C$$ to $$AB$$  
$$CF = \dfrac{12}{5}, AF = \dfrac{9}{5}   $$
在 $$ 	riangle CDF$$ 中
$$CD^2=CF^2+FD^2 = ( \dfrac{12}{5})^2+( \dfrac{9}{5}-x)^2$$
Since $$	riangle CED$$ , $$ 	riangle BCD $$ are similar
$$CD^2 = BD\cdotp DE  = (5-x)(5-x-y)$$ 
$$ ( \dfrac{12}{5})^2+( \dfrac{9}{5}-x)^2= (5-x)(5-x-y)$$
$$y = \dfrac{80-32x}{25-5x}  $$
$$(0 < x< \dfrac{5}{2} ) $$

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