Geometry Challenge: Find $\angle BCD$ in Convex Quadrilateral

[anemone]

Let $ABCD$ be a convex quadrilateral such that $AB=BC,\,AC=BD,\,\angle CBD=20^{\circ},\,\angle ABD=80^{\circ}$. Find $\angle BCD.$

 

[anemone]

Let $ABCD$ be a convex quadrilateral such that $AB=BC,\,AC=BD,\,\angle CBD=20^{\circ},\,\angle ABD=80^{\circ}$. Find $\angle BCD.$

Hint:

Spoiler

Draw a line parallel to $BD$ from $C$ and see where that leads you...

 

[anemone]

Hint:

Spoiler

Draw a line parallel to $BD$ from $C$ and see where that leads you...

Follow up hint:

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Construct an equilateral $\triangle ACE$ where $CE$ is the line parallel to $BD$.

 

[Albert1]

Let $ABCD$ be a convex quadrilateral such that $AB=BC,\,AC=BD,\,\angle CBD=20^{\circ},\,\angle ABD=80^{\circ}$. Find $\angle BCD.$

Spoiler

$130^o$

 

[MarkFL]

Spoiler

$130^o$

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Please provide how you obtained your solution. :)

 

[Albert1]

Spoiler

Please provide how you obtained your solution. :)

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step 1 :construct $\triangle ABC$ ,it is fixed , for $AB=BC$ , and $\angle ABC=20^o+80^o=100^o$
$\therefore \angle A=\angle C=40^o$
step 2: find point $D'$ on $AC$, and $\angle CBD'=20^o$
(step 1 and 2 can be done easily)
now point D must be on $BD'$ ray
step 3: construct circle $B$ with center $B$ and $BD=AC$
here point $D$ is the intersection of circle $B$ and $BD'$ ray
so $\angle BCD=\angle BCA+\angle ACD=40+90=130^o$

 

[anemone]

Spoiler

step 1 :construct $\triangle ABC$ ,it is fixed , for $AB=BC$ , and $\angle ABC=20^o+80^o=100^o$
$\therefore \angle A=\angle C=40^o$
step 2: find point $D'$ on $AC$, and $\angle CBD'=20^o$
(step 1 and 2 can be done easily)
now point D must be on $BD'$ ray
step 3: construct circle $B$ with center $B$ and $BD=AC$
here point $D$ is the intersection of circle $B$ and $BD'$ ray
so $\angle BCD=\angle BCA+\angle ACD=40+90=130^o$

Hi Albert!

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When you mentioned of constructing a circle that centered at $B$, I am wondering if $r=BA\stackrel{\text{or}}{=}BD$?

 

[Albert1]

Hi Albert!

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When you mentioned of constructing a circle that centered at $B$, I am wondering if $r=BA\stackrel{\text{or}}{=}BD$?

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$r=BD=AC$

 

[anemone]

Spoiler

$r=BD=AC$

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Okay, thanks for your reply...but Albert, I have constructed the diagram and drew everything up according to suggestion, but, I failed to see why $\angle ACD=90^\circ$.

 

[Albert1]

Spoiler

Okay, thanks for your reply...but Albert, I have constructed the diagram and drew everything up according to suggestion, but, I failed to see why $\angle ACD=90^\circ$.

Spoiler

View attachment 5532
$\angle ACD=90^o$

 

[anemone]

Spoiler

$\angle ACD=90^o$

Oh okay...thanks for your diagram!

Btw, the solution below is from a smart mathematician:

Spoiler

View attachment 5537

Draw a line parallel to $BD$ from $C$, and then construct an equilateral $\triangle ACE$ where $CE$ is the line that parallel to $BD.$

Note that $\triangle AEB \equiv \triangle CEB$ and since $BC=CB,\,\angle BCE=\angle CBD,\,CE=AC=BD$ (given), we have $\triangle BCE \equiv \triangle CBD$.

$\therefore \angle BCD=\angle CBE=180^\circ-20^\circ-\dfrac{60^\circ}{2}=130^\circ$