angle chasing with convex ABCD, equailateral ABD, isosceles BCD, <C=90^o

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angle chasing with convex ABCD, equailateral ABD, isosceles BCD, <C=90^o

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Source: May Olympiad (Olimpiada de Mayo) 2009 L2

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#1 h Sep 11, 2018, 7:14 PM • 1 Y

Y by Adventure10

Let $ABCD$ be a convex quadrilateral such that the triangle $ABD$ is equilateral and the triangle $BCD$ is isosceles, with $\angle C = 90^o$ . If $E$ is the midpoint of the side $AD$ , determine the measure of the angle $\angle CED$ .

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#2 h Sep 12, 2018, 12:06 AM • 2 Y

Y by Adventure10, Mango247

Coordinate bash! Let $B=(-1,0), D=(1,0)$ , then $C=(0,-1), A=(0, \sqrt{3}), E=(\frac{1}{2},\frac{\sqrt{3}}{2})$ . Then the cosine of the angle CED is equal to $\frac{CE^2+ED^2-CD^2}{2*CE*ED}$ by LoC, and computing the lengths using the Pythagorean theorem and other stuff leads us to obtain the value of the cosine as $\frac{1+\sqrt{3}}{2(2+\sqrt{3})}$ , so just take the arccosine of this number and you will get the answer.

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#3 h Sep 12, 2018, 5:56 AM • 1 Y

Y by Adventure10

$BCDE$ is cyclic, thus $\angle CED=\angle CBD=45^\circ$ .

Best regards,
sunken rock

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#4 h Sep 12, 2018, 7:44 AM • 2 Y

Y by Adventure10, Mango247

duck_master wrote:

... the value of the cosine as $\frac{1+\sqrt{3}}{2(2+\sqrt{3})}$ , so just take the arccosine of this number and you will get the answer.

There is a typo here. The correct is: $\cos x = \frac{{\sqrt 3 + 1}}{{2\sqrt {2 + \sqrt 3 } }} = \frac{{\sqrt 3 + 1}}{{2\sqrt {\frac{{{{(\sqrt 3 + 1)}^2}}}{2}} }} = \frac{{\sqrt 2 }}{2}...$ $\boxed{x=45^0}$

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#5 h Sep 12, 2018, 5:59 PM • 2 Y

Y by Adventure10, Mango247

Alternative solution without coordinates:
Let $F$ be the midpoint of $AB$ . Let $G$ be the intersection of the lines $AC$ and $EF$ . Let $H$ be the intersection of the lines $AC$ and $BD$ .
$AC=AH+HC=\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3}+1}{2}$

$GC=AC-AG=\frac{\sqrt{3}+1}{2}-\frac{\sqrt{3}}{4}=\frac{2\sqrt{3}-\sqrt{3}+2}{4}=\frac{\sqrt{3}+2}{4}$

Using the Pythagorean Theorem:
$EC=\sqrt{EG^2+GC^2}=\sqrt{\left(\frac{1}{4}\right)^2+\left(\frac{\sqrt{3}+2}{4}\right)^2}=\sqrt{\frac{1}{16}+\frac{3+4\sqrt{3}+4}{16}}=\frac{\sqrt{2+\sqrt{3}}}{2}=\frac{\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{2}}}{2}=\frac{\frac{\sqrt{3}+1}{\sqrt{2}}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$
Using the Law of Cosines in $\triangle EDC$ :
$DC^2=ED^2+EC^2-2\cdot ED\cdot EC \cos{\angle CED}$ $\left(\frac{\sqrt{2}}{2}\right)^2=\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)^2+2\cdot \frac{1}{2} \cdot \frac{\sqrt{6}+\sqrt{2}}{4} \cos{\angle CED}$ $\frac{1}{2}=\frac{1}{4}+\frac{2+\sqrt{3}}{4}-\frac{\sqrt{2}+\sqrt{6}}{4} \cos{\angle CED}$ $\frac{1}{2}-\frac{1}{4}-\frac{2+\sqrt{3}}{4}=-\frac{\sqrt{2}+\sqrt{6}}{4} \cos{\angle CED}$ $-\frac{1+\sqrt{3}}{4}=-\frac{\sqrt{2}+\sqrt{6}}{4} \cos{\angle CED}$ $\frac{1+\sqrt{3}}{4} \cdot \frac{4}{\sqrt{2}+\sqrt{6}}=\cos{\angle CED}$ $\frac{1+\sqrt{3}}{\sqrt{2}(1+\sqrt{3})}=\cos{\angle CED}$ $\frac{1}{\sqrt{2}}=\cos{\angle CED}$ $\boxed{\angle CED=45}$

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#6 h Jun 10, 2024, 10:40 AM

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look for cyclic quad

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