In the discussion about calculating angle $\angle ADC$ in convex quadrilateral $ABCD$, the angles given are $\angle CBD=10^{\circ}$, $\angle CAD=20^{\circ}$, $\angle ABD=40^{\circ}$, and $\angle BAC=50^{\circ}$. A participant pointed out that the initial diagram did not represent a convex quadrilateral correctly, suggesting an alternative configuration. The trigonometric approach used in the solution was nearly correct, but the final calculation needed adjustment. The corrected conclusion is that $\angle ADC$ measures $100^{\circ}$. The discussion emphasizes the importance of accurate geometric representation in solving angle problems.
#1
anemone
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In a convex quadrilateral $ABCD$ we are given that $\angle CBD=10^{\circ},\,\angle CAD=20^{\circ},\,\angle ABD=40^{\circ},\, \angle BAC=50^{\circ}$.
Let $B$ and $D$ be points on the $x$-axis and extend both the straight lines $BD$ and $AC$ such that they meet at $P$, where $\angle APB=90^{\circ}$ and $BP=k$.
\begin{tikzpicture}[auto]
\draw[thick, ->] (-1,0) -- (10,0) node[anchor = north west] {x};
\draw[thick, ->] (0,-9) -- (0,1) node[anchor = south east] {y};
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=right:P] (P) at (8,0);
\coordinate[label=left: D] (D) at (5,0);
\coordinate[label=right:C] (C) at (8,-3);
\coordinate[label=right:A] (A) at (8,-8);
\draw (B) to (P);
\draw (B) to (A);
\draw (D) to (A);
\draw (B) to (C);
\draw (A) to (P);
\draw (C) to (D);
\draw [dashed] (D) -- (3.6, 1.4);
\node (1) at (1.5,-0.2) {$10^{\circ}$};
\node (2) at (1.5,-0.9) {$30^{\circ}$};
\node (3) at (7,-6.3) {$30^{\circ}$};
\node (4) at (7.7,-6) {$20^{\circ}$};
\draw (P) rectangle +(-0.5, -0.5);
\end{tikzpicture}
This gives $\angle CDP=180^{\circ}-150^{\circ}=30^{\circ}$.
But $\angle ACP=30^{\circ}+40^{\circ}=70^{\circ}$, we get $\angle ADC=70^{\circ}-30^{\circ}=40^{\circ}$.
#3
kaliprasad
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Re: Find \$\angle ADC\$
anemone said:
My solution (I just solved it):
Let $B$ and $D$ be points on the $x$-axis and extend both the straight lines $BD$ and $AC$ such that they meet at $P$, where $\angle APB=90^{\circ}$ and $BP=k$.
\begin{tikzpicture}[auto]
\draw[thick, ->] (-1,0) -- (10,0) node[anchor = north west] {x};
\draw[thick, ->] (0,-9) -- (0,1) node[anchor = south east] {y};
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=right:P] (P) at (8,0);
\coordinate[label=left: D] (D) at (5,0);
\coordinate[label=right:C] (C) at (8,-3);
\coordinate[label=right:A] (A) at (8,-8);
\draw (B) to (P);
\draw (B) to (A);
\draw (D) to (A);
\draw (B) to (C);
\draw (A) to (P);
\draw (C) to (D);
\draw [dashed] (D) -- (3.6, 1.4);
\node (1) at (1.5,-0.2) {$10^{\circ}$};
\node (2) at (1.5,-0.9) {$30^{\circ}$};
\node (3) at (7,-6.3) {$30^{\circ}$};
\node (4) at (7.7,-6) {$20^{\circ}$};
\draw (P) rectangle +(-0.5, -0.5);
\end{tikzpicture}
This gives $\angle CDP=180^{\circ}-150^{\circ}=30^{\circ}$.
But $\angle ACP=30^{\circ}+40^{\circ}=70^{\circ}$, we get $\angle ADC=70^{\circ}-30^{\circ}=40^{\circ}$.
How do we know that AC is perpendicular to BD
#4
anemone
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Re: Find \$\angle ADC\$
kaliprasad said:
How do we know that AC is perpendicular to BD
Hi kaliprasad!
Notice that $\angle A=50^{\circ}$ and $\angle B=40^{\circ}$, if we extend the lines $AC$ and $BD$, to form a triangle, we see that we will get a right-angled triangle.
#5
Opalg
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Re: Find \$\angle ADC\$
anemone said:
My solution (I just solved it):
Let $B$ and $D$ be points on the $x$-axis and extend both the straight lines $BD$ and $AC$ such that they meet at $P$, where $\angle APB=90^{\circ}$ and $BP=k$.
\begin{tikzpicture}[auto]
\draw[thick, ->] (-1,0) -- (10,0) node[anchor = north west] {x};
\draw[thick, ->] (0,-9) -- (0,1) node[anchor = south east] {y};
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=right:P] (P) at (8,0);
\coordinate[label=left: D] (D) at (5,0);
\coordinate[label=right:C] (C) at (8,-3);
\coordinate[label=right:A] (A) at (8,-8);
\draw (B) to (P);
\draw (B) to (A);
\draw (D) to (A);
\draw (B) to (C);
\draw (A) to (P);
\draw (C) to (D);
\draw [dashed] (D) -- (3.6, 1.4);
\node (1) at (1.5,-0.2) {$10^{\circ}$};
\node (2) at (1.5,-0.9) {$30^{\circ}$};
\node (3) at (7,-6.3) {$30^{\circ}$};
\node (4) at (7.7,-6) {$20^{\circ}$};
\draw (P) rectangle +(-0.5, -0.5);
\end{tikzpicture}
This gives $\angle CDP=180^{\circ}-150^{\circ}=30^{\circ}$.
But $\angle ACP=30^{\circ}+40^{\circ}=70^{\circ}$, we get $\angle ADC=70^{\circ}-30^{\circ}=40^{\circ}$.
[sp]Very nice solution, anemone! My only criticism is that the quadrilateral $ABCD$ is supposed to be convex. If the vertices $A,B,C,D$ are to be taken in that order then your quadrilateral is not convex. I think that the diagram should look more like this:
[TIKZ] [scale=0.75]
\coordinate[label=left:A] (A) at (-10,0);
\coordinate[label=right:B] (B) at (0,12);
\coordinate[label=right:C] (C) at (2,0);
\coordinate[label=below right: D] (D) at (0,-3.5);
\draw (A) -- (B) -- (C) -- (D) -- (A) -- (C);
\draw (B) -- (D);
\node at (0.7,10.5) {$\leftarrow 10^{\circ}$};
\node at (-8,-0.3) {$20^{\circ}$};
\node at (-0.5,10.5) {$40^{\circ}$};
\node at (-8.5,0.5) {$50^{\circ}$};
\draw (0,0) rectangle +(-0.5, 0.5);
[/TIKZ]
Your trigonometric argument then works almost exactly as before, but the final result is that $\angle ADC$ is then $70^{\circ} + 30^{\circ}$ instead of $70^{\circ}-30^{\circ}$. So I think that the answer should be $100^\circ$.[/sp]
#6
anemone
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Re: Find \$\angle ADC\$
I initially struggled as I didn't know for sure if the vertices $A,B,C,D$ are to be taken in that order or not...after I saw there was no reply to this challenge, I decided to go with my hunch that it is okay they are not ordered in alphabetical manner. Now, the more I read it, the more I think you are right, so, thanks so much to Opalg for pointing it out...
Here is the amended solution (with much more neater trigonometric argument(Sun)):
[TIKZ] [scale=0.75]
\coordinate[label=left:A] (A) at (-10,0);
\coordinate[label=right:B] (B) at (0,12);
\coordinate[label=right:C] (C) at (2,0);
\coordinate[label=below right: D] (D) at (0,-3.5);
\coordinate[label=below right:P] (P) at (0,0);
\draw (A) -- (B) -- (C) -- (D) -- (A) -- (C);
\draw (B) -- (D);
\node at (0.7,10.5) {$\leftarrow 10^{\circ}$};
\node at (-8,-0.3) {$20^{\circ}$};
\node at (-0.5,10.5) {$40^{\circ}$};
\node at (-8.5,0.5) {$50^{\circ}$};
\draw (0,0) rectangle +(-0.5, 0.5);
[/TIKZ]Let $A$ be the origin point in the Cartesian plan, $AC$ and $BD$ meet at $P$ and $AP=k$.
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