Question

In the figure, point D lies on the side of AC in \triangle ABC, AC=25 , AB=35. Point E and F are movable on the side of AB (point F is on the left of point E) and \angle EDF=\angle A

If DF\perp AB , find the length of AE

Let x=AE , y=AF, determine the function y in terms of x and the domain of the function;

Connect CE, if \triangle DEC and \triangle ADF are similar, find the value of x

Collected in the board: Triangle

Steven Zheng posted 1 month ago


Answer

\because \tan A = \dfrac{4}{3}, AD =5,DF\perp AB

\therefore AF = 3, DF = 4

\because \angle EDF=\angle A

\therefore \triangle ADF \sim \triangle DEF

EF = \dfrac{4}{3}\cdotp DF = \dfrac{16}{3}

AE = AF+EF = 3+\dfrac{16}{3} =\dfrac{25}{3}


A second way to determine the length of AE

\triangle ADF \sim \triangle ADE

\dfrac{AE}{AD} = \dfrac{AD}{AF}

AE = AD^2\cdotp AF = \dfrac{25}{3}

\because \angle EDF=\angle A

\therefore \triangle DEF \sim \triangle AED

\dfrac{AE}{DE}=\dfrac{DE}{EF}

DE^2 = x(x-y)
(1)

In right \triangle DEG ,

DE^2 = DG^2+EG^2=4^2+(AE-AG)^2 =16+(x-3)^2
(2)
(1) and (2) yields the following equation

x(x-y) = 16+(x-3)^2

y = \dfrac{6x-25}{x} = 6-\dfrac{25}{x}

when y = 0, x = \dfrac{25}{6}

the domain of x is \dfrac{25}{6}\leq x\leq 35

Steven Zheng posted 1 month ago

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