如图，ABCD 是平行四边形。对角线 AC 和 BD 交于点 E。 $\overrightarrow{AF} = 2p$ 且 $\overrightarrow{AB} = 6p + 4q$ 。 F 是 AD 上的一点，使得 $\dfrac{AF}{AD} = \dfrac{1}{3}$ 。 G 是 AE 的中点。

((a))
用 p 和 q 尽可能简单地表示 $\overrightarrow{DE}$ 。
((b))
用 p 和 q 尽可能简单地表示 $\overrightarrow{AG}$ 。
((c))
证明 $\overrightarrow{BG} = -3p - 3q$ 。
((d))
解释为什么 B、G、F 三点共线。
((e))
计算：
三角形 ABF 的面积

平行四边形 ABCD 的面积

向量的加法与减法平行四边形与梯形面积向量数乘有理数比组合图形周长与面积最简比 English

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完整解法

6((a))

1. Find $\overrightarrow{AD}$ from the given ratio on side $AD$.

Since $F$ lies on $AD$ and $\dfrac{AF}{AD}=\dfrac13$ ,
$\overrightarrow{AF}=\frac13\overrightarrow{AD}.$
Given $\overrightarrow{AF}=2p$ ,
$\overrightarrow{AD}=3(2p)=6p.$

2. Use the diagonal property of a parallelogram to find $\overrightarrow{DE}$.

In parallelogram $ABCD$ ,
$\overrightarrow{DB}=\overrightarrow{DA}+\overrightarrow{AB}.$
Since $\overrightarrow{DA}=-\overrightarrow{AD}=-6p$ and $\overrightarrow{AB}=6p+4q$ ,
$\overrightarrow{DB}=-6p+(6p+4q)=4q.$
The diagonals of a parallelogram bisect each other, so $E$ is the midpoint of $DB$ .
Hence
$\overrightarrow{DE}=\frac12\overrightarrow{DB}=\frac12(4q)=2q.$

答案

$\overrightarrow{DE}=2q$

6((b))

1. Find $\overrightarrow{AD}$ using the given ratio $\dfrac{AF}{AD}=\dfrac{1}{3}$.

Since $\dfrac{AF}{AD}=\dfrac{1}{3}$ and $\overrightarrow{AF}=2p$ ,
$\overrightarrow{AD}=3\overrightarrow{AF}=3(2p)=6p.$

2. Find $\overrightarrow{AG}$ using the diagonal $AC$ and the midpoint properties of $E$ and $G$.

In parallelogram $ABCD$ ,
$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}.$
So
$\overrightarrow{AC}=(6p+4q)+6p=12p+4q.$
Since diagonals of a parallelogram bisect each other, $E$ is the midpoint of $AC$ :
$\overrightarrow{AE}=\frac{1}{2}\overrightarrow{AC}=\frac{1}{2}(12p+4q)=6p+2q.$
Since $G$ is the midpoint of $AE$ ,
$\overrightarrow{AG}=\frac{1}{2}\overrightarrow{AE}=\frac{1}{2}(6p+2q)=3p+q.$

答案

$\overrightarrow{AG}=3p+q$

6((c))

1. Express $\overrightarrow{BG}$ using a route through point $A$.

$\overrightarrow{BG}=\overrightarrow{BA}+\overrightarrow{AG}$
Since $\overrightarrow{AB}=6p+4q$ ,
$\overrightarrow{BA}=-(6p+4q).$

2. Substitute the known vector $\overrightarrow{AG}$ and simplify.

From part (b),
$\overrightarrow{AG}=3p+q.$
Therefore,
$\overrightarrow{BG}=-(6p+4q)+(3p+q)$
$=-6p-4q+3p+q$
$=-3p-3q.$

答案

$\overrightarrow{BG}=-3p-3q$ shown.

6((d))

1. Find a second vector from $B$ to a point on the suspected straight line.

$\overrightarrow{BF}=\overrightarrow{BA}+\overrightarrow{AF}$

$=-(6p+4q)+2p$

$=-4p-4q$

2. Compare $\overrightarrow{BF}$ with the known vector $\overrightarrow{BG}$ to prove collinearity.

From part (c),

$\overrightarrow{BG}=-3p-3q$ .

But

$\overrightarrow{BF}=-4p-4q$ .

$\overrightarrow{BG}=\dfrac{3}{4}\overrightarrow{BF}$ .

Since $\overrightarrow{BG}$ is a scalar multiple of $\overrightarrow{BF}$ , $BG$ is parallel to $BF$ . As both vectors start from $B$ , the points $B$ , $G$ and $F$ lie on a straight line.

答案

Since $\overrightarrow{BG}=\dfrac{1}{2}\overrightarrow{BF}$ , the vectors $\overrightarrow{BG}$ and $\overrightarrow{BF}$ are parallel and share the same starting point $B$ . Therefore, $B$ , $G$ and $F$ lie on a straight line.

6((e))

1. Relate the height of triangle $ABF$ to the height of parallelogram $ABCD$.

Since $\dfrac{AF}{AD}=\dfrac{1}{3}$ and $F$ lies on $AD$ , the perpendicular distance of $F$ from $AB$ is $\dfrac{1}{3}$ of the perpendicular height of parallelogram $ABCD$ from base $AB$ .

2. Compare the area of triangle $ABF$ with the area of parallelogram $ABCD$.

Let the base $AB$ be $b$ and the height of parallelogram $ABCD$ be $h$ .

$\text{Area of parallelogram }ABCD=bh$

$\text{Area of triangle }ABF=\frac{1}{2}\times b\times \frac{h}{3}=\frac{bh}{6}$

Hence,

$\frac{\text{area of triangle }ABF}{\text{area of parallelogram }ABCD}=\frac{\frac{bh}{6}}{bh}=\frac{1}{6}.$

答案

$\dfrac{\text{area of triangle }ABF}{\text{area of parallelogram }ABCD}=\dfrac{1}{6}$

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如图，ABCD 是平行四边形。 对角线 AC 和 BD 交于点 E。 AF→=2p\overrightarrow{AF} = 2pAF=2p 且 AB→=6p+4q\overrightarrow{AB} = 6p + 4qAB=6p+4q。 F 是 AD 上的一点，使得 AFAD=13\dfrac{AF}{AD} = \dfrac{1}{3}ADAF​=31​。 G 是 AE 的中点。