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

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

Uniteasy Admin · Accepted Answer

$$\because    	an A = \dfrac{4}{3}, AD =5,DF\perp AB $$
$$	herefore  AF = 3, DF = 4 $$
$$\because  \angle EDF=\angle A$$
$$	herefore   	riangle ADF \sim 	riangle DEF  $$
$$EF = \dfrac{4}{3}\cdotp DF = \dfrac{16}{3}    $$
$$AE = AF+EF = 3+\dfrac{16}{3} =\dfrac{25}{3}  $$

A second way to determine the length of AE
$$	riangle ADF \sim 	riangle ADE$$
$$\dfrac{AE}{AD}   = \dfrac{AD}{AF}  $$
$$AE = AD^2\cdotp AF = \dfrac{25}{3}   $$ $$\because  \angle EDF=\angle A$$
$$	herefore  	riangle  DEF \sim 	riangle AED $$
$$\dfrac{AE}{DE}=\dfrac{DE}{EF}    $$
$$DE^2 = x(x-y)$$n

In right $$	riangle DEG $$,
$$DE^2 = DG^2+EG^2=4^2+(AE-AG)^2 =16+(x-3)^2 $$n
(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 $$ \dfrac{25}{6}\leq x\leq 35  $$

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