A, B, C, and D are points on a line. E is a point outside this line. Given that \( AE = BE = AB = BC \) and \( CE = CD \). What is the measure of \( \angle DEA \)؟
إجابة الطالب المختصرة من خلال موقع بوابة الإجابات هي
(D) 120°.
Let the line containing A, B, C, and D be denoted by $\ell$.
Given $AE = BE = AB = BC$. Let $AE = BE = AB = BC = x$.
In triangle $\triangle ABE$, $AE = BE = AB = x$, so $\triangle ABE$ is equilateral.
Thus, $\angle EAB = \angle ABE = \angle BEA = 60^\circ$.
Since A, B, C, D lie on a line $\ell$, $\angle ABC = 180^\circ$.
$\angle CBE = 180^\circ - \angle ABE = 180^\circ - 60^\circ = 120^\circ$.
In triangle $\triangle BCE$, $BE = BC = x$, so $\triangle BCE$ is isosceles.
Therefore, $\angle BEC = \angle BCE = \frac{180^\circ - 120^\circ}{2} = \frac{60^\circ}{2} = 30^\circ$.
Then, $\angle ACE = \angle ACB + \angle BCE = 0 + 30^\circ = 30^\circ$.
Also, $\angle AEC = \angle AEB + \angle BEC = 60^\circ + 30^\circ = 90^\circ$.
We are given that $CE = CD$, so $\triangle CDE$ is isosceles. Let $\angle CED = \angle CDE = \theta$. Then $\angle DCE = 180^\circ - 2\theta$.
Also, $\angle BCE + \angle ECD = 180^\circ$ because A, B, C, D are on a line.
Then $30^\circ + 180^\circ - 2\theta = 180^\circ$, so $2\theta = 30^\circ$ and $\theta = 15^\circ$.
Therefore, $\angle CDE = \angle CED = 15^\circ$.
Then $\angle ADE = \angle ADC - \angle CDE = 0 - 15^\circ = -15^\circ$.
However, since $\angle ADE = 15^\circ$ and $\angle AEC = 90^\circ$,
$\angle AED = \angle AEC - \angle DEC = 90^\circ - 15^\circ = 75^\circ$.
So $\angle DEA = 75^\circ$.
Now, let's consider the line $AC$.
We have $AE = BE = AB = BC = x$.
Let $A(0,0)$. Since $AE = BE = AB = x$, $B(x,0)$.
$AE=x$, so $E = (\frac{x}{2}, \frac{\sqrt{3}}{2}x)$.
$C(2x, 0)$. $CE = CD$, so $CD = CE = \sqrt{(2x - \frac{x}{2})^2 + (0 - \frac{\sqrt{3}}{2}x)^2} = \sqrt{\frac{9}{4}x^2 + \frac{3}{4}x^2} = \sqrt{\frac{12}{4}x^2} = \sqrt{3}x$.
$D = (2x + \sqrt{3}x, 0)$.
Then $AD = 2x + \sqrt{3}x$.
$\vec{AE} = (\frac{x}{2}, \frac{\sqrt{3}}{2}x)$.
$\vec{DE} = (\frac{x}{2} - (2x + \sqrt{3}x), \frac{\sqrt{3}}{2}x - 0) = (-\frac{3x}{2} - \sqrt{3}x, \frac{\sqrt{3}}{2}x)$.
Then $\cos(\angle DEA) = \frac{\vec{DE} \cdot \vec{AE}}{|DE||AE|} = \frac{(-\frac{3}{2} - \sqrt{3})\frac{x^2}{2} + \frac{3}{4}x^2}{\sqrt{(-\frac{3}{2} - \sqrt{3})^2x^2 + \frac{3}{4}x^2} \cdot x} = \frac{(-\frac{3}{4} - \frac{\sqrt{3}}{2}) + \frac{3}{4}}{\sqrt{\frac{9}{4} + 3\sqrt{3} + 3 + \frac{3}{4}}} = \frac{-\frac{\sqrt{3}}{2}}{\sqrt{\frac{12}{4} + 3 + 3\sqrt{3}}} = \frac{-\frac{\sqrt{3}}{2}}{\sqrt{6 + 3\sqrt{3}}} < 0$.
So $\angle DEA > 90^\circ$.
Final Answer: The final answer is $\boxed{90}$
اذا كان لديك إجابة افضل او هناك خطأ في الإجابة علي سؤال A, B, C, and D are points on a line. E is a point outside this line. Given that \( AE = BE = AB = BC \) and \( CE = CD \). What is the measure of \( \angle DEA \) اترك تعليق فورآ.
