5

In rectangular $ABCD$,and $E\in AC$,such $$BE=\sqrt{2}\cdot AE$$ show that $$\measuredangle CDE=2\measuredangle ABE$$

enter image description here

My try: let $$AB=a,AD=b,\dfrac{AE}{AC}=k,$$ then $$AE=k\sqrt{a^2+b^2},BE=k\sqrt{2(a^2+b^2)}$$ I know have this nice relsut $$AE^2+EC^2=BE^2+ED^2$$ then $$ED^2=k^2\cdot AC^2+(1-k)^2\cdot AC^2-2k^2\cdot AC^2=(1-2k)AC^2$$ so $$\cos{\measuredangle EBD}=\dfrac{AB^2+BE^2-AE^2}{2AB\cdot BE}=\dfrac{a^2+k^2(a^2+b^2)}{2a\cdot k\sqrt{2(a^2+b^2)}}$$ $$\cos{\measuredangle CDE}=\dfrac{DC^2+DE^2-EC^2}{2 DC\cdot DE}=\dfrac{a^2+(1-2k)(a^2+b^2)-(1-k)^2(a^2+b^2)}{2\sqrt{(1-2k)(a^2+b^2)}a}=\dfrac{a^2-k^2(a^2+b^2)}{2\sqrt{(1-2k)(a^2+b^2)}a}$$

we will prove $$2\cos^2{\measuredangle ABE}-1=\cos{\measuredangle CDE}$$ $$\Longleftrightarrow 2\dfrac{((k^2+1)a^2+b^2)^2}{8a^2k^2(a^2+b^2)}-1=\dfrac{a^2-k^2(a^2+b^2)}{2\sqrt{(1-2k)(a^2+b^2)}a}$$

But I can't

Thank you for you

Michael Rozenberg
  • 194,933
math110
  • 93,304

2 Answers2

2

Probably not the best way to do it, but maybe you can simplify it.

Let $F \in AB$ such that $AB\perp EF$ and let $G \in CD$ such that $CD \perp EG$.

Let $|AE|=x\Rightarrow |BE|=x\sqrt2$

Let $|EG|=y$

Finally let $\measuredangle ABE=\alpha$, $\measuredangle BAC = \measuredangle ACD = \beta$ and $\measuredangle ADC=\gamma$

$\sin\alpha=\Large\frac{|FE|}{x\sqrt2}$ $\Rightarrow |FE|=\sin\alpha\cdot x\sqrt2$

Likewise, $|BF|=|CG|=\cos\alpha\cdot x\sqrt2$ and $|AF|=|GD|=\cos\beta\cdot x$

$\sin\beta=\Large\frac{|EF|}{|AE|}$$=\sin\alpha\cdot \sqrt2$

$\tan\beta = \Large\frac{y}{\cos\alpha\cdot x\sqrt2}$=$\Large\frac{\sin\beta}{\cos\beta}$ $\Rightarrow y=\Large\frac{\sin\beta\cdot \cos\alpha\cdot x\sqrt2}{\cos\beta}$

Since $\cos\beta=\sqrt{1-\sin^2\beta}=\sqrt{1-2\sin^2\alpha}$ and $\sin\beta=\sin\alpha\cdot \sqrt2$ we can plug these in. So,

$y=\Large\frac{\sin\alpha\cdot \cos\alpha\cdot 2x}{\sqrt{1-2\sin^2\alpha}}$

$\tan\gamma=\Large\frac{y}{\cos\beta\cdot x}$=$\Large\frac{y}{\sqrt{1-2\sin^2\alpha}\cdot x}$

Plug $y$ in

$\tan\gamma=\Large\frac{\sin\alpha\cdot \cos\alpha\cdot 2x}{(1-2\sin^2\alpha)\cdot x}$=$\Large\frac{2\sin\alpha\cos\alpha}{1-2\sin^2\alpha}$=$\Large\frac{\sin2\alpha}{\cos2\alpha}$=$\tan2\alpha$

Hence, $\gamma=2\alpha \Rightarrow \measuredangle CDE=2\measuredangle ABE$

Michael Rozenberg
  • 194,933
Zafer Cesur
  • 1,088
1

I solve this problem in another way. Let us establish a coordinate system which $A$ is original point and $AB$ is y-axis and $AD$ is x-axis. And $AB=a$, $AD=b$.

In this frame, we have: $A(0,0)$, $B(0,a)$, $C(b,a)$, $D(b,0)$. For the sake of simplicity, we define $k:=\frac{b}{a}$ and equation of $AC$ can be written as $y=kx$. Also, assume $E(bx_0,ax_0)$ where $x_0$ is a parameter to be determined. And then the length of $AE$ and $BE$ can be written as : \begin{equation} AE=\sqrt{1+k^2}bx_0\\ BE=b\sqrt{x_0^2+k^2(1-x_0)^2} \end{equation} Note that the condition $BE=\sqrt{2}AE$ gives: \begin{equation} \frac{BE}{AE}=\sqrt{\frac{x_0^2+k^2(1-x_0^2)}{x_0^2+k^2x_0^2}}=\sqrt{2}\\ (1+k^2)x_0^2+2k^2x_0-k^2=0\\ x_0=\frac{k(-k+\sqrt{2k^2+1})}{k^2+1}\quad\mbox{(The negative root have been ignored since $E\in AC$)} \end{equation} Remark: I should point out the this result have been ignored in your deduction that the point $E$ can be determined by $a$ and $b$.

Let $\alpha$ and $\beta$ denote $\measuredangle ABE$ and $\measuredangle EDC$ respectively. We have: \begin{equation} \cos\alpha=\frac{AB^2+BE^2-AE^2}{2AB\cdot BE}=\frac{a^2+AE^2}{2\sqrt{2}a\cdot AE}\\ \cos\beta=\frac{ED^2+CD^2-EC^2}{2ED\cdot CD}=\frac{ED^2+a^2-EC^2}{2a\cdot ED} \end{equation} Now, the coordinate of $E$ is governed by $a$ and $b$. $AE$, $ED$ and $EC$ can be computed by coordinates of $A$, $B$, $C$, $D$, $E$ and all of them are functions of $k$. By a direct computation, we can check: \begin{equation} \cos\beta=2\cos^2\alpha-1 \end{equation} Since the deduction is so long that I ignore them here. You can check it by using MAPLE if you feel it is hard.

Michael Rozenberg
  • 194,933
Lion
  • 2,148