图中 ABCDE 是一个正五边形。 BE 和 BD 是直线。

(a)
求以下各项的值（并给出理由）：
1. (i)
  角 ABE。
  
  答案 ___
2. (ii)
  角 EBD。
  
  答案 ___
(b)
解释为什么 BE 平行于 CD。

答案 ___
(c)
延长边 AE 和 CD 相交于点 X。
1. (i)
  计算角 DEX。
  
  答案 ___ °
2. (ii)
  解释为什么四边形 BDXE 是一个平行四边形。
  
  答案 ___

几何图形性质平行线与截线角的关系多边形内角和与外角和 English

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完整解法

16(a)(i)

1. Find the angle at $A$ in the regular pentagon.

Since $ABCDE$ is a regular pentagon, each interior angle is
$\frac{(5-2)\times 180^\circ}{5}=108^\circ.$

2. Use triangle $ABE$ to find $\angle ABE$.

In $\triangle ABE$ , $AB=AE$ because all sides of a regular pentagon are equal.

So $\triangle ABE$ is isosceles, and
$\angle ABE=\angle BEA.$
Hence
$\angle ABE=\frac{180^\circ-108^\circ}{2}=36^\circ.$

答案

$\angle ABE = 36^\circ$

16(a)(ii)

1. Find the angle between $BE$ and $BD$ at $B$.

In regular pentagon $ABCDE$ , $\angle ABC=108^\circ$ .

In $\triangle BCD$ , $BC=CD$ , so
$\angle CBD=\angle BDC=\dfrac{180^\circ-108^\circ}{2}=36^\circ$ .

Hence
$\angle ABD=108^\circ-36^\circ=72^\circ$ .

Using $\angle ABE=36^\circ$ from part (a)(i),
$\angle EBD=72^\circ-36^\circ=36^\circ.$

答案

$\angle EBD = 36^\circ$

16(b)

1. Compare a pair of alternate interior angles formed by transversal $BD$.

From part (a)(ii), $\angle EBD = 36^\circ$ .

In a regular pentagon, diagonal $BD$ makes $\angle BDC = 36^\circ$ by the same base-angle reasoning in isosceles triangle $BCD$ .

Hence $\angle EBD = \angle BDC$ .

Therefore, $BE \parallel CD$ since alternate interior angles are equal.

答案

$\angle EBD = \angle BDC = 36^\circ$ , so $BE \parallel CD$ since alternate interior angles are equal.

16(c)(i)

1. Use angles on a straight line at $E$ to find $\angle DEX$.

In a regular pentagon, each interior angle is $108^\circ$ , so $\angle AED = 108^\circ$ .

Since $AE$ is produced to $X$ , $AEX$ is a straight line.

$\angle DEX = 180^\circ - 108^\circ = 72^\circ$

答案

$72^\circ$

16(c)(ii)

1. Show that one pair of opposite sides is parallel.

Since $X$ lies on $CD$ produced, $D$ , $C$ and $X$ are collinear.

From part (b), $BE \parallel CD$ .

Therefore, $BE \parallel DX$ .

2. Show that the other pair of opposite sides is parallel.

At $D$ ,
$\angle BDE = \angle CDE - \angle BDC.$
Since $ABCDE$ is a regular pentagon, $\angle CDE = 108^\circ$ .

Also, $\angle BDC = 36^\circ$ , so
$\angle BDE = 108^\circ - 36^\circ = 72^\circ.$
From part (c)(i),
$\angle DEX = 72^\circ.$
Thus $\angle BDE = \angle DEX$ , so $BD \parallel EX$ .

3. Conclude using the definition/property of a parallelogram.

In quadrilateral $BDXE$ ,
$BE \parallel DX \quad \text{and} \quad BD \parallel EX.$
Since both pairs of opposite sides are parallel, $BDXE$ is a parallelogram.

答案

Since $X$ lies on $CD$ produced, $DX$ is the same straight line as $CD$ . From part (b), $BE \parallel CD$ , so $BE \parallel DX$ . Also, $\angle BDE = 108^\circ - 36^\circ = 72^\circ$ and from part (c)(i), $\angle DEX = 72^\circ$ . Hence $BD \parallel EX$ . Therefore, both pairs of opposite sides of $BDXE$ are parallel, so $BDXE$ is a parallelogram.

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