已知三角形 $ABC$ 中， $AC=8$ , $AB=\sqrt{41}$ , $BC$ 边上的高 $AG=5$ , $D$ 为线段 $AC$ 上的动点, 在 $BC$ 上截取 $CE=AD$ , 连接 $AE$ , $AC=8$ $0$ , 则 $AC=8$ $1$ 的最小值为

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Q. 已知三角形

ABC

中，

AC=8

AB=\sqrt{41}

BC

边上的高

AG=5

D

为线段

AC

上的动点, 在

BC

上截取

CE=AD

, 连接

AE

AC=8

0

, 则

AC=8

1

的最小值为

Use Triangle Area Formula: Use the triangle area formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ . Calculate the area of triangle ABC using base BC and height AG. $\text{Area}_{\text{ABC}} = \frac{1}{2} \times \text{BC} \times \text{AG}$
Calculate Area of Triangle ABC: We need to find the length of BC. We can use the Pythagorean theorem since $AG$ is the height and $AB$ is the hypotenuse of the right triangle $ABG$ . $\newline$ $BC^2 = AB^2 - AG^2$ $\newline$ $BC^2 = (\sqrt{41})^2 - 5^2$ $\newline$ $BC^2 = 41 - 25$ $\newline$ $BC^2 = 16$ $\newline$ $BC = \sqrt{16}$ $\newline$ $BC = 4$
Find Length of BC: Now we have the length of BC, we can find the area of triangle ABC. $\newline$ Area $_{ABC}$ = $\frac{1}{2} \times BC \times AG$ $\newline$ Area $_{ABC}$ = $\frac{1}{2} \times 4 \times 5$ $\newline$ Area $_{ABC}$ = $10$
Find Area of Triangle $ABC$ : Since $CE=AD$ and $D$ is a point on $AC$ , triangles $ADE$ and $CEB$ are similar by $AA$ similarity (both have a right angle and share angle $AEC$ ).
Triangles $ADE$ and $CEB$ are Similar: The sum $AE+BD$ is minimized when $AE$ and $BD$ are both at their minimum lengths. This occurs when $D$ is the midpoint of $AC$ , making $AD = DC = \frac{AC}{2}$ . $AD = DC = \frac{8}{2}$ $AD = DC = 4$
Find Length of AD: Since triangles ADE and CEB are similar, the ratio of their corresponding sides is equal. $\newline$ $\frac{AE}{AD} = \frac{BC}{CE}$ $\newline$ $\frac{AE}{4} = \frac{4}{4}$ $\newline$ $AE = 4$
Find Length of AE: Now we can find $BD$ using the Pythagorean theorem in triangle $BDC$ . $\newline$ $BD^2 = BC^2 + DC^2$ $\newline$ $BD^2 = 4^2 + 4^2$ $\newline$ $BD^2 = 16 + 16$ $\newline$ $BD^2 = 32$ $\newline$ $BD = \sqrt{32}$ $\newline$ $BD = 4\sqrt{2}$
Find Length of BD: Finally, we can find the minimum value of $AE+BD$ . $AE+BD = 4 + 4\sqrt{2}$