In the figure, is a \triangle 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 .

If CD\perp AB ,find the value of the length of BE

If \triangle CDE is an isosceles triangle, find the length of AD

If AD=x , BE=y , find function y in terms of x and the domain of the function.

Collected in the board: Triangle

Steven Zheng posted 1 month ago


In \triangle CED and \triangle BCD

\because \angle DCE = \angle B and they share the same angle \angle BDC

\therefore The two triangles are similar, and their corresponding sides are proportional.

\dfrac{DC}{BD} = \dfrac{ED}{DC}

DC^2 = BD\cdotp ED

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}


=\dfrac{16}{5} - \dfrac{9}{5} =\dfrac{7}{5}

Since \triangle CED \triangle ACD are similar, \triangle 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}

\triangle CDF

CD^2=CF^2+FD^2 = ( \dfrac{12}{5})^2+( \dfrac{9}{5}-x)^2

Since \triangle CED , \triangle 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} )

Steven Zheng posted 1 month ago

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