Mixing
Ratio of angles in a triangle
Clash Royale CLAN TAG#URR8PPP
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came across this question in a math contest and can't quite figure out an approach to the question.
Triangle $ABC$ has $AB=ACneq BC$ and $∠BAC ≤ 90º$. $P$ lies on $AC$, and $Q$ lies on $AB$ such that $AP = PQ = QB = B$C. Find the ratio of $∠ACB$ to $∠APQ$.
I have tried to establish some sort of congruency between the inner triangles but to no success. So far, all I have is the obvious result that:
$180º=4 times ∠ACB - ∠APQ$
Any tips or hints are greatly appreciated.
geometry contest-math
asked Aug 6 at 12:24
oscquito
1438
add a comment |Â
up vote
3
down vote
favorite
came across this question in a math contest and can't quite figure out an approach to the question.
Triangle $ABC$ has $AB=ACneq BC$ and $∠BAC ≤ 90º$. $P$ lies on $AC$, and $Q$ lies on $AB$ such that $AP = PQ = QB = B$C. Find the ratio of $∠ACB$ to $∠APQ$.
I have tried to establish some sort of congruency between the inner triangles but to no success. So far, all I have is the obvious result that:
$180º=4 times ∠ACB - ∠APQ$
Any tips or hints are greatly appreciated.
geometry contest-math
asked Aug 6 at 12:24
oscquito
1438
Have you noticed that $∠BQP=2times∠ACB$?
– Servaes
Aug 6 at 12:36
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
came across this question in a math contest and can't quite figure out an approach to the question.
Triangle $ABC$ has $AB=ACneq BC$ and $∠BAC ≤ 90º$. $P$ lies on $AC$, and $Q$ lies on $AB$ such that $AP = PQ = QB = B$C. Find the ratio of $∠ACB$ to $∠APQ$.
I have tried to establish some sort of congruency between the inner triangles but to no success. So far, all I have is the obvious result that:
$180º=4 times ∠ACB - ∠APQ$
Any tips or hints are greatly appreciated.
geometry contest-math
asked Aug 6 at 12:24
oscquito
1438
came across this question in a math contest and can't quite figure out an approach to the question.
Triangle $ABC$ has $AB=ACneq BC$ and $∠BAC ≤ 90º$. $P$ lies on $AC$, and $Q$ lies on $AB$ such that $AP = PQ = QB = B$C. Find the ratio of $∠ACB$ to $∠APQ$.
I have tried to establish some sort of congruency between the inner triangles but to no success. So far, all I have is the obvious result that:
$180º=4 times ∠ACB - ∠APQ$
Any tips or hints are greatly appreciated.
geometry contest-math
asked Aug 6 at 12:24
oscquito
1438
edited Aug 6 at 12:31
asked Aug 6 at 12:24
oscquito
1438
asked Aug 6 at 12:24
oscquito
1438
asked Aug 6 at 12:24
oscquito
1438
1438
Have you noticed that $∠BQP=2times∠ACB$?
– Servaes
Aug 6 at 12:36
add a comment |Â
Have you noticed that $∠BQP=2times∠ACB$?
– Servaes
Aug 6 at 12:36
Have you noticed that $∠BQP=2times∠ACB$?
– Servaes
Aug 6 at 12:36
Have you noticed that $∠BQP=2times∠ACB$?
– Servaes
Aug 6 at 12:36
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Let us set $AP=PQ=QB=BC=1$, the unit.
Let us denote by $x$ the angle in $A$. Then the sine theorem in $Delta APQ$ gives
$$
AQ = APcdotfracsin(180^circ-2x)sin x
=APcdotfrac sin(2x)sin x
=APcdot 2cos x
=2cos x
.
$$
The sides of $Delta ABC$ are thus $BC=1$ and $1+2cos x$, and the last again. The sine theorem gives again
$$
frac sin x1
=fracsin(90^circ-x/2)1+2cos x .
$$
This gives an equation in $x$ to be solved.
We rewrite it equivalently step for step:
$$
beginaligned
sin x &=fraccos(x/2)1+2cos x ,\
sin x(1+2cos x) &= cosfrac x2 ,\
sin x + sin 2x &=cosfrac x2 ,\
2sin frac3x2cos frac x2 &= cosfrac x2 ,\
2sin frac3x2 &= 1 ,\
sin frac3x2 &= frac 12 ,\
frac3x2 &= 30^circ ,\
x &= 20^circ .
endaligned
$$
Now we have the angles in the triangle(s).
$widehatAPQ$ has measure $180^circ-2x=140^circ$.
The angles at the base are both $80^circ$.
The problem asks now for a ration that makes no sense (geometrically). It is even misleading (in a contest, because many people would first suppose it is a beautiful ration, thus loosing 5 minutes or more).
Note: I will try to give also a synthetic proof.
answered Aug 6 at 13:00
dan_fulea
4,2171211
add a comment |Â
up vote
1
down vote
Clue to the Problem
Find a point $D$ such that $QD ||BC $ and $QD=BC$.
- $BCDQ$ is a rhombus;
$triangle APQ cong triangle PDC$;
$triangle PQD$ is an Equilateral triangle;
- $angle A=20^o$
- $cdots$
answered Aug 6 at 13:13
mengdie1982
2,972216
Knowing the answer, $20^circ$, it is natural that the above picture "will work". (It is obtained by gathering together two congruent isosceles triangles, $Delta ABC$, and $Delta CA?$, so that $ABC?$ is a parallelogram.) It remains to "make it work" by argument. Why are the two triangles $APQ$, $PDC$ congruent? (I.e. why is $Delta PDC$ isosceles?)
– dan_fulea
Aug 6 at 19:22
add a comment |Â
up vote
1
down vote
I promised a synthetic answer, there is a clue answer already posted in the mean time, so i am inserting an other construction.
The idea is to use the given construction "in a copied manner" on the same sheet of paper, then find special constellations of points that allow to deduce angles in the figure. (It is natural to put the same triangle "side by side".)
So let us consider the following situation...
We start with the triangle $Delta ABC$, and points $Pin AC$, $Qin AB$, so that we have the "snake of segments" $APQBC$, explicitly $$AP=PQ=QB=BC .$$ Let $D$ be the symmetric of $C$ w.r.t. the perpendicular bisector of the side $AB$, so $ABCD$ is isosceles, and $AB=AC=BD$. The "snake" $APQBC$ can then be seen in the triangles $Delta ABC$, $Delta ACB$, $Delta BAD$, $Delta BDA$ in different fashions,
$$APQBC equiv AP'SCB equiv BQRDA equiv BQ'P'AD ,$$
thus uniquely defining the points $P',Q',R,S$ as in the picture. Let $x$ be the angle in $A$ in $Delta ABC$.
Then the triangles $Delta APD$ and $Delta Q'BC$ are equilateral and $x=20^circ$.
Proof: The triangles $Delta PQ'B$ and $Delta APQ$ are isosceles, so their angles in $P,Q$ are also $x$. By symmetry w.r.t. the perpendicular bisector $(s)$ of $AB$ we have the correspondences $Aleftrightarrow B$, and $Cleftrightarrow D$, so by construction we also have the correspondences $P'leftrightarrow Q$, $Pleftrightarrow Q'$, $Rleftrightarrow S$. In particular, $PQ| AB$ (both being $perp s$). It follows that $AP'Q'P$ is a parallelogram with equal sides, thus a rhombus. In particular, $BP$ is the angle bisector of the angle $angle ABD$, so $$AP=PD .$$
Form this, $AP=PD=DA$, so $Delta APD$ is equilateral. (Same also for $Delta BQ'C$.)
The sum of the angles in $A$, and $D$ in the trapez $ABCD$ is $180^circ$, in the sum we consider twice an angle of $60^circ$ from $Delta APD$ and further $3x$. From here, we get $x=20^circ$.
$square$
answered Aug 6 at 19:17
dan_fulea
4,2171211
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let us set $AP=PQ=QB=BC=1$, the unit.
Let us denote by $x$ the angle in $A$. Then the sine theorem in $Delta APQ$ gives
$$
AQ = APcdotfracsin(180^circ-2x)sin x
=APcdotfrac sin(2x)sin x
=APcdot 2cos x
=2cos x
.
$$
The sides of $Delta ABC$ are thus $BC=1$ and $1+2cos x$, and the last again. The sine theorem gives again
$$
frac sin x1
=fracsin(90^circ-x/2)1+2cos x .
$$
This gives an equation in $x$ to be solved.
We rewrite it equivalently step for step:
$$
beginaligned
sin x &=fraccos(x/2)1+2cos x ,\
sin x(1+2cos x) &= cosfrac x2 ,\
sin x + sin 2x &=cosfrac x2 ,\
2sin frac3x2cos frac x2 &= cosfrac x2 ,\
2sin frac3x2 &= 1 ,\
sin frac3x2 &= frac 12 ,\
frac3x2 &= 30^circ ,\
x &= 20^circ .
endaligned
$$
Now we have the angles in the triangle(s).
$widehatAPQ$ has measure $180^circ-2x=140^circ$.
The angles at the base are both $80^circ$.
The problem asks now for a ration that makes no sense (geometrically). It is even misleading (in a contest, because many people would first suppose it is a beautiful ration, thus loosing 5 minutes or more).
Note: I will try to give also a synthetic proof.
answered Aug 6 at 13:00
dan_fulea
4,2171211
add a comment |Â
up vote
3
down vote
accepted
Let us set $AP=PQ=QB=BC=1$, the unit.
Let us denote by $x$ the angle in $A$. Then the sine theorem in $Delta APQ$ gives
$$
AQ = APcdotfracsin(180^circ-2x)sin x
=APcdotfrac sin(2x)sin x
=APcdot 2cos x
=2cos x
.
$$
The sides of $Delta ABC$ are thus $BC=1$ and $1+2cos x$, and the last again. The sine theorem gives again
$$
frac sin x1
=fracsin(90^circ-x/2)1+2cos x .
$$
This gives an equation in $x$ to be solved.
We rewrite it equivalently step for step:
$$
beginaligned
sin x &=fraccos(x/2)1+2cos x ,\
sin x(1+2cos x) &= cosfrac x2 ,\
sin x + sin 2x &=cosfrac x2 ,\
2sin frac3x2cos frac x2 &= cosfrac x2 ,\
2sin frac3x2 &= 1 ,\
sin frac3x2 &= frac 12 ,\
frac3x2 &= 30^circ ,\
x &= 20^circ .
endaligned
$$
Now we have the angles in the triangle(s).
$widehatAPQ$ has measure $180^circ-2x=140^circ$.
The angles at the base are both $80^circ$.
The problem asks now for a ration that makes no sense (geometrically). It is even misleading (in a contest, because many people would first suppose it is a beautiful ration, thus loosing 5 minutes or more).
Note: I will try to give also a synthetic proof.
answered Aug 6 at 13:00
dan_fulea
4,2171211
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let us set $AP=PQ=QB=BC=1$, the unit.
Let us denote by $x$ the angle in $A$. Then the sine theorem in $Delta APQ$ gives
$$
AQ = APcdotfracsin(180^circ-2x)sin x
=APcdotfrac sin(2x)sin x
=APcdot 2cos x
=2cos x
.
$$
The sides of $Delta ABC$ are thus $BC=1$ and $1+2cos x$, and the last again. The sine theorem gives again
$$
frac sin x1
=fracsin(90^circ-x/2)1+2cos x .
$$
This gives an equation in $x$ to be solved.
We rewrite it equivalently step for step:
$$
beginaligned
sin x &=fraccos(x/2)1+2cos x ,\
sin x(1+2cos x) &= cosfrac x2 ,\
sin x + sin 2x &=cosfrac x2 ,\
2sin frac3x2cos frac x2 &= cosfrac x2 ,\
2sin frac3x2 &= 1 ,\
sin frac3x2 &= frac 12 ,\
frac3x2 &= 30^circ ,\
x &= 20^circ .
endaligned
$$
Now we have the angles in the triangle(s).
$widehatAPQ$ has measure $180^circ-2x=140^circ$.
The angles at the base are both $80^circ$.
The problem asks now for a ration that makes no sense (geometrically). It is even misleading (in a contest, because many people would first suppose it is a beautiful ration, thus loosing 5 minutes or more).
Note: I will try to give also a synthetic proof.
answered Aug 6 at 13:00
dan_fulea
4,2171211
Let us set $AP=PQ=QB=BC=1$, the unit.
Let us denote by $x$ the angle in $A$. Then the sine theorem in $Delta APQ$ gives
$$
AQ = APcdotfracsin(180^circ-2x)sin x
=APcdotfrac sin(2x)sin x
=APcdot 2cos x
=2cos x
.
$$
The sides of $Delta ABC$ are thus $BC=1$ and $1+2cos x$, and the last again. The sine theorem gives again
$$
frac sin x1
=fracsin(90^circ-x/2)1+2cos x .
$$
This gives an equation in $x$ to be solved.
We rewrite it equivalently step for step:
$$
beginaligned
sin x &=fraccos(x/2)1+2cos x ,\
sin x(1+2cos x) &= cosfrac x2 ,\
sin x + sin 2x &=cosfrac x2 ,\
2sin frac3x2cos frac x2 &= cosfrac x2 ,\
2sin frac3x2 &= 1 ,\
sin frac3x2 &= frac 12 ,\
frac3x2 &= 30^circ ,\
x &= 20^circ .
endaligned
$$
Now we have the angles in the triangle(s).
$widehatAPQ$ has measure $180^circ-2x=140^circ$.
The angles at the base are both $80^circ$.
The problem asks now for a ration that makes no sense (geometrically). It is even misleading (in a contest, because many people would first suppose it is a beautiful ration, thus loosing 5 minutes or more).
Note: I will try to give also a synthetic proof.
answered Aug 6 at 13:00
dan_fulea
4,2171211
answered Aug 6 at 13:00
dan_fulea
4,2171211
answered Aug 6 at 13:00
dan_fulea
4,2171211
answered Aug 6 at 13:00
dan_fulea
4,2171211
4,2171211
add a comment |Â
add a comment |Â
up vote
1
down vote
Clue to the Problem
Find a point $D$ such that $QD ||BC $ and $QD=BC$.
- $BCDQ$ is a rhombus;
$triangle APQ cong triangle PDC$;
$triangle PQD$ is an Equilateral triangle;
- $angle A=20^o$
- $cdots$
answered Aug 6 at 13:13
mengdie1982
2,972216
Knowing the answer, $20^circ$, it is natural that the above picture "will work". (It is obtained by gathering together two congruent isosceles triangles, $Delta ABC$, and $Delta CA?$, so that $ABC?$ is a parallelogram.) It remains to "make it work" by argument. Why are the two triangles $APQ$, $PDC$ congruent? (I.e. why is $Delta PDC$ isosceles?)
– dan_fulea
Aug 6 at 19:22
add a comment |Â
up vote
1
down vote
Clue to the Problem
Find a point $D$ such that $QD ||BC $ and $QD=BC$.
- $BCDQ$ is a rhombus;
$triangle APQ cong triangle PDC$;
$triangle PQD$ is an Equilateral triangle;
- $angle A=20^o$
- $cdots$
answered Aug 6 at 13:13
mengdie1982
2,972216
Knowing the answer, $20^circ$, it is natural that the above picture "will work". (It is obtained by gathering together two congruent isosceles triangles, $Delta ABC$, and $Delta CA?$, so that $ABC?$ is a parallelogram.) It remains to "make it work" by argument. Why are the two triangles $APQ$, $PDC$ congruent? (I.e. why is $Delta PDC$ isosceles?)
– dan_fulea
Aug 6 at 19:22
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Clue to the Problem
Find a point $D$ such that $QD ||BC $ and $QD=BC$.
- $BCDQ$ is a rhombus;
$triangle APQ cong triangle PDC$;
$triangle PQD$ is an Equilateral triangle;
- $angle A=20^o$
- $cdots$
answered Aug 6 at 13:13
mengdie1982
2,972216
Clue to the Problem
Find a point $D$ such that $QD ||BC $ and $QD=BC$.
- $BCDQ$ is a rhombus;
$triangle APQ cong triangle PDC$;
$triangle PQD$ is an Equilateral triangle;
- $angle A=20^o$
- $cdots$
answered Aug 6 at 13:13
mengdie1982
2,972216
answered Aug 6 at 13:13
mengdie1982
2,972216
answered Aug 6 at 13:13
mengdie1982
2,972216
answered Aug 6 at 13:13
mengdie1982
2,972216
2,972216
Knowing the answer, $20^circ$, it is natural that the above picture "will work". (It is obtained by gathering together two congruent isosceles triangles, $Delta ABC$, and $Delta CA?$, so that $ABC?$ is a parallelogram.) It remains to "make it work" by argument. Why are the two triangles $APQ$, $PDC$ congruent? (I.e. why is $Delta PDC$ isosceles?)
– dan_fulea
Aug 6 at 19:22
add a comment |Â
Knowing the answer, $20^circ$, it is natural that the above picture "will work". (It is obtained by gathering together two congruent isosceles triangles, $Delta ABC$, and $Delta CA?$, so that $ABC?$ is a parallelogram.) It remains to "make it work" by argument. Why are the two triangles $APQ$, $PDC$ congruent? (I.e. why is $Delta PDC$ isosceles?)
– dan_fulea
Aug 6 at 19:22
Knowing the answer, $20^circ$, it is natural that the above picture "will work". (It is obtained by gathering together two congruent isosceles triangles, $Delta ABC$, and $Delta CA?$, so that $ABC?$ is a parallelogram.) It remains to "make it work" by argument. Why are the two triangles $APQ$, $PDC$ congruent? (I.e. why is $Delta PDC$ isosceles?)
– dan_fulea
Aug 6 at 19:22
Knowing the answer, $20^circ$, it is natural that the above picture "will work". (It is obtained by gathering together two congruent isosceles triangles, $Delta ABC$, and $Delta CA?$, so that $ABC?$ is a parallelogram.) It remains to "make it work" by argument. Why are the two triangles $APQ$, $PDC$ congruent? (I.e. why is $Delta PDC$ isosceles?)
– dan_fulea
Aug 6 at 19:22
add a comment |Â
up vote
1
down vote
I promised a synthetic answer, there is a clue answer already posted in the mean time, so i am inserting an other construction.
The idea is to use the given construction "in a copied manner" on the same sheet of paper, then find special constellations of points that allow to deduce angles in the figure. (It is natural to put the same triangle "side by side".)
So let us consider the following situation...
We start with the triangle $Delta ABC$, and points $Pin AC$, $Qin AB$, so that we have the "snake of segments" $APQBC$, explicitly $$AP=PQ=QB=BC .$$ Let $D$ be the symmetric of $C$ w.r.t. the perpendicular bisector of the side $AB$, so $ABCD$ is isosceles, and $AB=AC=BD$. The "snake" $APQBC$ can then be seen in the triangles $Delta ABC$, $Delta ACB$, $Delta BAD$, $Delta BDA$ in different fashions,
$$APQBC equiv AP'SCB equiv BQRDA equiv BQ'P'AD ,$$
thus uniquely defining the points $P',Q',R,S$ as in the picture. Let $x$ be the angle in $A$ in $Delta ABC$.
Then the triangles $Delta APD$ and $Delta Q'BC$ are equilateral and $x=20^circ$.
Proof: The triangles $Delta PQ'B$ and $Delta APQ$ are isosceles, so their angles in $P,Q$ are also $x$. By symmetry w.r.t. the perpendicular bisector $(s)$ of $AB$ we have the correspondences $Aleftrightarrow B$, and $Cleftrightarrow D$, so by construction we also have the correspondences $P'leftrightarrow Q$, $Pleftrightarrow Q'$, $Rleftrightarrow S$. In particular, $PQ| AB$ (both being $perp s$). It follows that $AP'Q'P$ is a parallelogram with equal sides, thus a rhombus. In particular, $BP$ is the angle bisector of the angle $angle ABD$, so $$AP=PD .$$
Form this, $AP=PD=DA$, so $Delta APD$ is equilateral. (Same also for $Delta BQ'C$.)
The sum of the angles in $A$, and $D$ in the trapez $ABCD$ is $180^circ$, in the sum we consider twice an angle of $60^circ$ from $Delta APD$ and further $3x$. From here, we get $x=20^circ$.
$square$
answered Aug 6 at 19:17
dan_fulea
4,2171211
add a comment |Â
up vote
1
down vote
I promised a synthetic answer, there is a clue answer already posted in the mean time, so i am inserting an other construction.
The idea is to use the given construction "in a copied manner" on the same sheet of paper, then find special constellations of points that allow to deduce angles in the figure. (It is natural to put the same triangle "side by side".)
So let us consider the following situation...
We start with the triangle $Delta ABC$, and points $Pin AC$, $Qin AB$, so that we have the "snake of segments" $APQBC$, explicitly $$AP=PQ=QB=BC .$$ Let $D$ be the symmetric of $C$ w.r.t. the perpendicular bisector of the side $AB$, so $ABCD$ is isosceles, and $AB=AC=BD$. The "snake" $APQBC$ can then be seen in the triangles $Delta ABC$, $Delta ACB$, $Delta BAD$, $Delta BDA$ in different fashions,
$$APQBC equiv AP'SCB equiv BQRDA equiv BQ'P'AD ,$$
thus uniquely defining the points $P',Q',R,S$ as in the picture. Let $x$ be the angle in $A$ in $Delta ABC$.
Then the triangles $Delta APD$ and $Delta Q'BC$ are equilateral and $x=20^circ$.
Proof: The triangles $Delta PQ'B$ and $Delta APQ$ are isosceles, so their angles in $P,Q$ are also $x$. By symmetry w.r.t. the perpendicular bisector $(s)$ of $AB$ we have the correspondences $Aleftrightarrow B$, and $Cleftrightarrow D$, so by construction we also have the correspondences $P'leftrightarrow Q$, $Pleftrightarrow Q'$, $Rleftrightarrow S$. In particular, $PQ| AB$ (both being $perp s$). It follows that $AP'Q'P$ is a parallelogram with equal sides, thus a rhombus. In particular, $BP$ is the angle bisector of the angle $angle ABD$, so $$AP=PD .$$
Form this, $AP=PD=DA$, so $Delta APD$ is equilateral. (Same also for $Delta BQ'C$.)
The sum of the angles in $A$, and $D$ in the trapez $ABCD$ is $180^circ$, in the sum we consider twice an angle of $60^circ$ from $Delta APD$ and further $3x$. From here, we get $x=20^circ$.
$square$
answered Aug 6 at 19:17
dan_fulea
4,2171211
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I promised a synthetic answer, there is a clue answer already posted in the mean time, so i am inserting an other construction.
The idea is to use the given construction "in a copied manner" on the same sheet of paper, then find special constellations of points that allow to deduce angles in the figure. (It is natural to put the same triangle "side by side".)
So let us consider the following situation...
We start with the triangle $Delta ABC$, and points $Pin AC$, $Qin AB$, so that we have the "snake of segments" $APQBC$, explicitly $$AP=PQ=QB=BC .$$ Let $D$ be the symmetric of $C$ w.r.t. the perpendicular bisector of the side $AB$, so $ABCD$ is isosceles, and $AB=AC=BD$. The "snake" $APQBC$ can then be seen in the triangles $Delta ABC$, $Delta ACB$, $Delta BAD$, $Delta BDA$ in different fashions,
$$APQBC equiv AP'SCB equiv BQRDA equiv BQ'P'AD ,$$
thus uniquely defining the points $P',Q',R,S$ as in the picture. Let $x$ be the angle in $A$ in $Delta ABC$.
Then the triangles $Delta APD$ and $Delta Q'BC$ are equilateral and $x=20^circ$.
Proof: The triangles $Delta PQ'B$ and $Delta APQ$ are isosceles, so their angles in $P,Q$ are also $x$. By symmetry w.r.t. the perpendicular bisector $(s)$ of $AB$ we have the correspondences $Aleftrightarrow B$, and $Cleftrightarrow D$, so by construction we also have the correspondences $P'leftrightarrow Q$, $Pleftrightarrow Q'$, $Rleftrightarrow S$. In particular, $PQ| AB$ (both being $perp s$). It follows that $AP'Q'P$ is a parallelogram with equal sides, thus a rhombus. In particular, $BP$ is the angle bisector of the angle $angle ABD$, so $$AP=PD .$$
Form this, $AP=PD=DA$, so $Delta APD$ is equilateral. (Same also for $Delta BQ'C$.)
The sum of the angles in $A$, and $D$ in the trapez $ABCD$ is $180^circ$, in the sum we consider twice an angle of $60^circ$ from $Delta APD$ and further $3x$. From here, we get $x=20^circ$.
$square$
answered Aug 6 at 19:17
dan_fulea
4,2171211
I promised a synthetic answer, there is a clue answer already posted in the mean time, so i am inserting an other construction.
The idea is to use the given construction "in a copied manner" on the same sheet of paper, then find special constellations of points that allow to deduce angles in the figure. (It is natural to put the same triangle "side by side".)
So let us consider the following situation...
We start with the triangle $Delta ABC$, and points $Pin AC$, $Qin AB$, so that we have the "snake of segments" $APQBC$, explicitly $$AP=PQ=QB=BC .$$ Let $D$ be the symmetric of $C$ w.r.t. the perpendicular bisector of the side $AB$, so $ABCD$ is isosceles, and $AB=AC=BD$. The "snake" $APQBC$ can then be seen in the triangles $Delta ABC$, $Delta ACB$, $Delta BAD$, $Delta BDA$ in different fashions,
$$APQBC equiv AP'SCB equiv BQRDA equiv BQ'P'AD ,$$
thus uniquely defining the points $P',Q',R,S$ as in the picture. Let $x$ be the angle in $A$ in $Delta ABC$.
Then the triangles $Delta APD$ and $Delta Q'BC$ are equilateral and $x=20^circ$.
Proof: The triangles $Delta PQ'B$ and $Delta APQ$ are isosceles, so their angles in $P,Q$ are also $x$. By symmetry w.r.t. the perpendicular bisector $(s)$ of $AB$ we have the correspondences $Aleftrightarrow B$, and $Cleftrightarrow D$, so by construction we also have the correspondences $P'leftrightarrow Q$, $Pleftrightarrow Q'$, $Rleftrightarrow S$. In particular, $PQ| AB$ (both being $perp s$). It follows that $AP'Q'P$ is a parallelogram with equal sides, thus a rhombus. In particular, $BP$ is the angle bisector of the angle $angle ABD$, so $$AP=PD .$$
Form this, $AP=PD=DA$, so $Delta APD$ is equilateral. (Same also for $Delta BQ'C$.)
The sum of the angles in $A$, and $D$ in the trapez $ABCD$ is $180^circ$, in the sum we consider twice an angle of $60^circ$ from $Delta APD$ and further $3x$. From here, we get $x=20^circ$.
$square$
answered Aug 6 at 19:17
dan_fulea
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answered Aug 6 at 19:17
dan_fulea
4,2171211
answered Aug 6 at 19:17
dan_fulea
4,2171211
answered Aug 6 at 19:17
dan_fulea
4,2171211
4,2171211
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Have you noticed that $∠BQP=2times∠ACB$?
– Servaes
Aug 6 at 12:36