Finding angle with pure Geometry.perimeter of square inscribed in the triagleTricky pure geometry proofSolid Geometry. Finding an angleGeometry- finding the measure of the angleTwo tangent circles inscribed in a rectangle (Compute the area)Geometry: Finding an angle of a trapezoidFinding angle and radius. Geometry/Trig applicationGeometry (Finding angle $x$)Foundations and Fundamental Concepts of Mathematics, Chapter Problems (1.1.3)Find the two missing angles in a quadrilateral
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Finding angle with pure Geometry.
perimeter of square inscribed in the triagleTricky pure geometry proofSolid Geometry. Finding an angleGeometry- finding the measure of the angleTwo tangent circles inscribed in a rectangle (Compute the area)Geometry: Finding an angle of a trapezoidFinding angle and radius. Geometry/Trig applicationGeometry (Finding angle $x$)Foundations and Fundamental Concepts of Mathematics, Chapter Problems (1.1.3)Find the two missing angles in a quadrilateral
$begingroup$
Each side of a square ABCD has a length of 1 unit. Points P and Q belong to AB and DA respectively. The perimeter of triangle APQ is 2 units. What will be the angle of PCQ.
I was able to do this with simple trignometry and found it to be 45 degrees but the book I am reading doesn't yet talked about trignometry or similarity of triangle so I want rather pure geometric proof which doesn't include trigonometry or similarity of triangle concepts.
geometry euclidean-geometry
asked 2 hours ago
Keshav SharmaKeshav Sharma
1286
$endgroup$
add a comment |
$begingroup$
Each side of a square ABCD has a length of 1 unit. Points P and Q belong to AB and DA respectively. The perimeter of triangle APQ is 2 units. What will be the angle of PCQ.
I was able to do this with simple trignometry and found it to be 45 degrees but the book I am reading doesn't yet talked about trignometry or similarity of triangle so I want rather pure geometric proof which doesn't include trigonometry or similarity of triangle concepts.
geometry euclidean-geometry
asked 2 hours ago
Keshav SharmaKeshav Sharma
1286
$endgroup$
$begingroup$
Let $AQ=x,AP=y.$ Then you can find that $2x+2y=xy+2,$ which has infinitely many solutions for $0 < x,y leq 1.$
$endgroup$
– Dbchatto67
2 hours ago
add a comment |
$begingroup$
Each side of a square ABCD has a length of 1 unit. Points P and Q belong to AB and DA respectively. The perimeter of triangle APQ is 2 units. What will be the angle of PCQ.
I was able to do this with simple trignometry and found it to be 45 degrees but the book I am reading doesn't yet talked about trignometry or similarity of triangle so I want rather pure geometric proof which doesn't include trigonometry or similarity of triangle concepts.
geometry euclidean-geometry
asked 2 hours ago
Keshav SharmaKeshav Sharma
1286
$endgroup$
Each side of a square ABCD has a length of 1 unit. Points P and Q belong to AB and DA respectively. The perimeter of triangle APQ is 2 units. What will be the angle of PCQ.
I was able to do this with simple trignometry and found it to be 45 degrees but the book I am reading doesn't yet talked about trignometry or similarity of triangle so I want rather pure geometric proof which doesn't include trigonometry or similarity of triangle concepts.
geometry euclidean-geometry
geometry euclidean-geometry
asked 2 hours ago
Keshav SharmaKeshav Sharma
1286
asked 2 hours ago
Keshav SharmaKeshav Sharma
1286
asked 2 hours ago
Keshav SharmaKeshav Sharma
1286
asked 2 hours ago
Keshav SharmaKeshav Sharma
1286
asked 2 hours ago
Keshav SharmaKeshav Sharma
1286
1286
$begingroup$
Let $AQ=x,AP=y.$ Then you can find that $2x+2y=xy+2,$ which has infinitely many solutions for $0 < x,y leq 1.$
$endgroup$
– Dbchatto67
2 hours ago
add a comment |
$begingroup$
Let $AQ=x,AP=y.$ Then you can find that $2x+2y=xy+2,$ which has infinitely many solutions for $0 < x,y leq 1.$
$endgroup$
– Dbchatto67
2 hours ago
$begingroup$
Let $AQ=x,AP=y.$ Then you can find that $2x+2y=xy+2,$ which has infinitely many solutions for $0 < x,y leq 1.$
$endgroup$
– Dbchatto67
2 hours ago
$begingroup$
Let $AQ=x,AP=y.$ Then you can find that $2x+2y=xy+2,$ which has infinitely many solutions for $0 < x,y leq 1.$
$endgroup$
– Dbchatto67
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
(I apologize for replacing points $PQ$ with $EF$, I'm too tired to draw the picture again :)
Draw circle with center $E$ and radius $EB$ and circle with center $F$ and radius $FD$. These circles intersect segment $EF$ at points $G',G''$ These points are identical! Why?
Because perimeter of triangle $AEF$ is 2 which is $AB+AD=AE+EB+AF+FD=AE+EG'+AF+FG''$. It means that $EG'+FG''=EF$ and therefore $G'equiv G''equiv Gin EF$.
So these two circles touch at point $G$ as shown in the picture. Power of point $C$ with respect to both circles is equal $(CB=CD)$ and therefore it has to be on the radical axis of the circles. These circles touch at point $G$ and their radical axis is defined by the common tangent at point $G$. Becuase of that $CG$ must be tangent to both circles and, at the same time, $CGbot EF$.
The rest is trivial: you can easily show that triangles $FCD$ and $FCG$ are congruent. The same is true for triangles $ECG$ and $ECB$. Because of that:
$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$
answered 1 hour ago
OldboyOldboy
9,37911138
$endgroup$
add a comment |
$begingroup$
Rotate $Q$ for $90^circ$ around $C$ in to new point $E$. Then $P,B,E$ are collinear and $PB = PQ$. So triangles $CQP$ and $CEP$ are congruent by (sss), so $$angle QCP = angle ECP = 45^circ$$
answered 56 mins ago
Maria MazurMaria Mazur
49.9k1361124
$endgroup$
$begingroup$
Could you provide a picture?
$endgroup$
– Dr. Mathva
43 mins ago
$begingroup$
@Dr.Mathva It is not a difficult to do it your self. Just rotate the $F$ (instead of $Q$) in oldboys picture.
$endgroup$
– Maria Mazur
41 mins ago
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
10 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
(I apologize for replacing points $PQ$ with $EF$, I'm too tired to draw the picture again :)
Draw circle with center $E$ and radius $EB$ and circle with center $F$ and radius $FD$. These circles intersect segment $EF$ at points $G',G''$ These points are identical! Why?
Because perimeter of triangle $AEF$ is 2 which is $AB+AD=AE+EB+AF+FD=AE+EG'+AF+FG''$. It means that $EG'+FG''=EF$ and therefore $G'equiv G''equiv Gin EF$.
So these two circles touch at point $G$ as shown in the picture. Power of point $C$ with respect to both circles is equal $(CB=CD)$ and therefore it has to be on the radical axis of the circles. These circles touch at point $G$ and their radical axis is defined by the common tangent at point $G$. Becuase of that $CG$ must be tangent to both circles and, at the same time, $CGbot EF$.
The rest is trivial: you can easily show that triangles $FCD$ and $FCG$ are congruent. The same is true for triangles $ECG$ and $ECB$. Because of that:
$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$
answered 1 hour ago
OldboyOldboy
9,37911138
$endgroup$
add a comment |
$begingroup$
(I apologize for replacing points $PQ$ with $EF$, I'm too tired to draw the picture again :)
Draw circle with center $E$ and radius $EB$ and circle with center $F$ and radius $FD$. These circles intersect segment $EF$ at points $G',G''$ These points are identical! Why?
Because perimeter of triangle $AEF$ is 2 which is $AB+AD=AE+EB+AF+FD=AE+EG'+AF+FG''$. It means that $EG'+FG''=EF$ and therefore $G'equiv G''equiv Gin EF$.
So these two circles touch at point $G$ as shown in the picture. Power of point $C$ with respect to both circles is equal $(CB=CD)$ and therefore it has to be on the radical axis of the circles. These circles touch at point $G$ and their radical axis is defined by the common tangent at point $G$. Becuase of that $CG$ must be tangent to both circles and, at the same time, $CGbot EF$.
The rest is trivial: you can easily show that triangles $FCD$ and $FCG$ are congruent. The same is true for triangles $ECG$ and $ECB$. Because of that:
$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$
answered 1 hour ago
OldboyOldboy
9,37911138
$endgroup$
add a comment |
$begingroup$
(I apologize for replacing points $PQ$ with $EF$, I'm too tired to draw the picture again :)
Draw circle with center $E$ and radius $EB$ and circle with center $F$ and radius $FD$. These circles intersect segment $EF$ at points $G',G''$ These points are identical! Why?
Because perimeter of triangle $AEF$ is 2 which is $AB+AD=AE+EB+AF+FD=AE+EG'+AF+FG''$. It means that $EG'+FG''=EF$ and therefore $G'equiv G''equiv Gin EF$.
So these two circles touch at point $G$ as shown in the picture. Power of point $C$ with respect to both circles is equal $(CB=CD)$ and therefore it has to be on the radical axis of the circles. These circles touch at point $G$ and their radical axis is defined by the common tangent at point $G$. Becuase of that $CG$ must be tangent to both circles and, at the same time, $CGbot EF$.
The rest is trivial: you can easily show that triangles $FCD$ and $FCG$ are congruent. The same is true for triangles $ECG$ and $ECB$. Because of that:
$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$
answered 1 hour ago
OldboyOldboy
9,37911138
$endgroup$
(I apologize for replacing points $PQ$ with $EF$, I'm too tired to draw the picture again :)
Draw circle with center $E$ and radius $EB$ and circle with center $F$ and radius $FD$. These circles intersect segment $EF$ at points $G',G''$ These points are identical! Why?
Because perimeter of triangle $AEF$ is 2 which is $AB+AD=AE+EB+AF+FD=AE+EG'+AF+FG''$. It means that $EG'+FG''=EF$ and therefore $G'equiv G''equiv Gin EF$.
So these two circles touch at point $G$ as shown in the picture. Power of point $C$ with respect to both circles is equal $(CB=CD)$ and therefore it has to be on the radical axis of the circles. These circles touch at point $G$ and their radical axis is defined by the common tangent at point $G$. Becuase of that $CG$ must be tangent to both circles and, at the same time, $CGbot EF$.
The rest is trivial: you can easily show that triangles $FCD$ and $FCG$ are congruent. The same is true for triangles $ECG$ and $ECB$. Because of that:
$$angle ECF=frac12angle BCG+frac 12angle GCD=45^circ$$
answered 1 hour ago
OldboyOldboy
9,37911138
answered 1 hour ago
OldboyOldboy
9,37911138
answered 1 hour ago
OldboyOldboy
9,37911138
answered 1 hour ago
OldboyOldboy
9,37911138
9,37911138
add a comment |
add a comment |
$begingroup$
Rotate $Q$ for $90^circ$ around $C$ in to new point $E$. Then $P,B,E$ are collinear and $PB = PQ$. So triangles $CQP$ and $CEP$ are congruent by (sss), so $$angle QCP = angle ECP = 45^circ$$
answered 56 mins ago
Maria MazurMaria Mazur
49.9k1361124
$endgroup$
$begingroup$
Could you provide a picture?
$endgroup$
– Dr. Mathva
43 mins ago
$begingroup$
@Dr.Mathva It is not a difficult to do it your self. Just rotate the $F$ (instead of $Q$) in oldboys picture.
$endgroup$
– Maria Mazur
41 mins ago
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
10 mins ago
add a comment |
$begingroup$
Rotate $Q$ for $90^circ$ around $C$ in to new point $E$. Then $P,B,E$ are collinear and $PB = PQ$. So triangles $CQP$ and $CEP$ are congruent by (sss), so $$angle QCP = angle ECP = 45^circ$$
answered 56 mins ago
Maria MazurMaria Mazur
49.9k1361124
$endgroup$
$begingroup$
Could you provide a picture?
$endgroup$
– Dr. Mathva
43 mins ago
$begingroup$
@Dr.Mathva It is not a difficult to do it your self. Just rotate the $F$ (instead of $Q$) in oldboys picture.
$endgroup$
– Maria Mazur
41 mins ago
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
10 mins ago
add a comment |
$begingroup$
Rotate $Q$ for $90^circ$ around $C$ in to new point $E$. Then $P,B,E$ are collinear and $PB = PQ$. So triangles $CQP$ and $CEP$ are congruent by (sss), so $$angle QCP = angle ECP = 45^circ$$
answered 56 mins ago
Maria MazurMaria Mazur
49.9k1361124
$endgroup$
Rotate $Q$ for $90^circ$ around $C$ in to new point $E$. Then $P,B,E$ are collinear and $PB = PQ$. So triangles $CQP$ and $CEP$ are congruent by (sss), so $$angle QCP = angle ECP = 45^circ$$
answered 56 mins ago
Maria MazurMaria Mazur
49.9k1361124
edited 47 mins ago
answered 56 mins ago
Maria MazurMaria Mazur
49.9k1361124
answered 56 mins ago
Maria MazurMaria Mazur
49.9k1361124
answered 56 mins ago
Maria MazurMaria Mazur
49.9k1361124
49.9k1361124
$begingroup$
Could you provide a picture?
$endgroup$
– Dr. Mathva
43 mins ago
$begingroup$
@Dr.Mathva It is not a difficult to do it your self. Just rotate the $F$ (instead of $Q$) in oldboys picture.
$endgroup$
– Maria Mazur
41 mins ago
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
10 mins ago
add a comment |
$begingroup$
Could you provide a picture?
$endgroup$
– Dr. Mathva
43 mins ago
$begingroup$
@Dr.Mathva It is not a difficult to do it your self. Just rotate the $F$ (instead of $Q$) in oldboys picture.
$endgroup$
– Maria Mazur
41 mins ago
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
10 mins ago
$begingroup$
Could you provide a picture?
$endgroup$
– Dr. Mathva
43 mins ago
$begingroup$
Could you provide a picture?
$endgroup$
– Dr. Mathva
43 mins ago
$begingroup$
@Dr.Mathva It is not a difficult to do it your self. Just rotate the $F$ (instead of $Q$) in oldboys picture.
$endgroup$
– Maria Mazur
41 mins ago
$begingroup$
@Dr.Mathva It is not a difficult to do it your self. Just rotate the $F$ (instead of $Q$) in oldboys picture.
$endgroup$
– Maria Mazur
41 mins ago
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
10 mins ago
$begingroup$
You're right... And how can $PB=PQ$? Isn't that a particular case?
$endgroup$
– Dr. Mathva
10 mins ago
add a comment |
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$begingroup$
Let $AQ=x,AP=y.$ Then you can find that $2x+2y=xy+2,$ which has infinitely many solutions for $0 < x,y leq 1.$
$endgroup$
– Dbchatto67
2 hours ago