- #1
anemone
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Let $ABCD$ be a convex quadrilateral such that $AB=BC,\,AC=BD,\,\angle CBD=20^{\circ},\,\angle ABD=80^{\circ}$. Find $\angle BCD.$
Geometry Challenge: Find $\angle BCD$ in Convex Quadrilateral
In summary, a convex quadrilateral is a four-sided polygon with all interior angles less than 180 degrees. The formula for finding $\angle BCD$ in a convex quadrilateral is $\angle BCD = 180 - (\angle ABC + \angle ADC)$, and the sum of the interior angles in a convex quadrilateral is always 360 degrees. The measure of $\angle BCD$ cannot exceed 180 degrees in a convex quadrilateral, with two special cases being when $\angle ABC$ and $\angle ADC$ are either supplementary or equal.
- #2
anemone
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anemone said:
Hint:
Draw a line parallel to $BD$ from $C$ and see where that leads you...
- #3
anemone
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anemone said:
Follow up hint:
Construct an equilateral $\triangle ACE$ where $CE$ is the line parallel to $BD$.
- #4
Albert1
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anemone said:
$130^o$
Last edited:
- #5
MarkFL
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Albert said:
Please provide how you obtained your solution. :)
- #6
Albert1
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MarkFL said:
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 $\angle BCD=\angle BCA+\angle ACD=40+90=130^o$
$\therefore \angle A=\angle C=40^o$
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 $\angle BCD=\angle BCA+\angle ACD=40+90=130^o$
Last edited:
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anemone
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Albert said:
Hi Albert!
When you mentioned of constructing a circle that centered at $B$, I am wondering if $r=BA\stackrel{\text{or}}{=}BD$?
- #8
Albert1
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anemone said:
$r=BD=AC$
- #9
anemone
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Albert said:
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 $\angle ACD=90^\circ$.
- #10
Albert1
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anemone said:
View attachment 5532
$\angle ACD=90^o$
$\angle ACD=90^o$
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- #11
anemone
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Albert said:
Oh okay...thanks for your diagram!
Btw, the solution below is from a smart mathematician:
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 $\triangle BCE \equiv \triangle CBD$.
$\therefore \angle BCD=\angle CBE=180^\circ-20^\circ-\dfrac{60^\circ}{2}=130^\circ$
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 $\triangle BCE \equiv \triangle CBD$.
$\therefore \angle BCD=\angle CBE=180^\circ-20^\circ-\dfrac{60^\circ}{2}=130^\circ$
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Related to Geometry Challenge: Find $\angle BCD$ in Convex Quadrilateral
What is a convex quadrilateral?
A convex quadrilateral is a four-sided polygon with all interior angles less than 180 degrees. This means that all of its vertices "point outwards" and the diagonal line segments connecting its vertices lie entirely within the shape.
How do I find $\angle BCD$ in a convex quadrilateral?
To find $\angle BCD$ in a convex quadrilateral, you can use the formula: $\angle BCD = 180 - (\angle ABC + \angle ADC)$, where $\angle ABC$ and $\angle ADC$ are the adjacent angles to $\angle BCD$.
What is the sum of the interior angles in a convex quadrilateral?
The sum of the interior angles in a convex quadrilateral is always 360 degrees. This is because a convex quadrilateral can be divided into two triangles, each with interior angles summing to 180 degrees, so the total sum is 360 degrees.
Can the measure of $\angle BCD$ be greater than 180 degrees in a convex quadrilateral?
No, the measure of $\angle BCD$ cannot be greater than 180 degrees in a convex quadrilateral. As mentioned earlier, a convex quadrilateral has all interior angles less than 180 degrees, so $\angle BCD$ cannot exceed this limit.
Are there any special cases for finding $\angle BCD$ in a convex quadrilateral?
Yes, there are two special cases for finding $\angle BCD$ in a convex quadrilateral. The first case is when $\angle ABC$ and $\angle ADC$ are supplementary angles (add up to 180 degrees), in which case $\angle BCD$ would be 0 degrees. The second case is when $\angle ABC$ and $\angle ADC$ are equal, in which case $\angle BCD$ would be 90 degrees.
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