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Difference between revisions of "Congruent (geometry)"

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'''Congruency''' is a property of multiple geometric figures.
 +
 
== Intuitive Definition ==
 
== Intuitive Definition ==
  
−
Two [[geometry | geometric]] [[figure]]s are '''congruent''' if one of them can be turned and/or flipped and placed exactly on top of the other, with all parts lining up perfectly with no parts on either figure left over.  In plain language, two objects are congruent if they have the same size and shape.
+
Two [[geometry | geometric]] [[figure]]s are congruent if one of them can be turned and/or flipped and placed exactly on top of the other, with all parts lining up perfectly with no parts on either figure left over.  In plain language, two objects are congruent if they have the same size and shape.
  
 +
[[Image:reflection.PNG|thumb|right|100px|100px|A collection of isometries.]]
  
 
== Technical Definition ==
 
== Technical Definition ==
  
−
Two [[geometry | geometric]] objects are '''congruent''' if one can be transformed into the other by an [[isometry]], such as a [[translation]], [[rotation]], [[reflection]] or some combination thereof.   
+
Two [[geometry | geometric]] objects are '''congruent''' if one can be transformed into the other by an [[isometry]], such as a [[translation]], [[rotation]], [[reflection]] or some combination thereof.
 +
 
 +
== Axioms ==
 +
 
 +
*If <math>A</math>, <math>B</math> are two points on a straight line <math>a</math>, and if <math>A'</math> is a point upon the same or another straight line <math>a'</math>, then, upon a given side of <math>A'</math> on the straight line <math>a'</math>, we can always find one and only one point <math>B'</math> so that the segment <math>AB</math> (or <math>BA</math>) is congruent to the segment <math>A'B'</math>. We indicate this relation by writing
 +
<cmath>\overline{AB}\cong\overline{A'B'}.</cmath>
 +
Every segment is congruent to itself; that is, we always have
 +
<cmath>\overline{AB}\cong\overline{AB}.</cmath>
 +
 
 +
*If a segment <math>\overline{AB}</math> is congruent to the segment <math>\overline{A'B'}</math> and also to the segment <math>\overline{A''B''}</math>, then the segment <math>\overline{A'B'}</math> is congruent to the segment <math>\overline{A''B''}</math>; that is, if <math>\overline{AB}\cong\overline{A'B}</math> and <math>\overline{AB}\cong\overline{A''B''}</math>, then <math>\overline{A'B'}\cong\overline{A''B''}</math>.
 +
 
 +
*Let <math>\overline{AB}</math> and <math>\overline{BC}</math> be two segments of a straight line <math>a</math> which have no points in common aside from the point <math>B</math>, and, furthermore, let <math>\overline{A'B'}</math> and <math>\overline{B'C'}</math> be two segments of the same or of another straight line <math>a'</math> having, likewise, no point other than <math>B'</math> in common.  Then, if <math>\overline{AB}\cong\overline{A'B'}</math> and <math>\overline{BC}\cong\overline{B'C'}</math>, we have <math>\overline{AC}\cong\overline{A'C'}</math>.
 +
 
 +
*Let an angle <math>(h,k)</math> be given in the plane <math>\alpha</math> and let a straight line <math>a'</math> be given in a plane <math>\alpha'</math>. Suppose also that, in the plane <math>\alpha</math>, a definite side of the straight line <math>a'</math> be assigned. Denote by <math>h'</math> a half-ray of the straight line <math>a'</math> emanating from a point <math>O'</math> of this line. Then in the plane <math>\alpha'</math> there is one and only one half-ray <math>k'</math> such that the angle <math>(h,k)</math>, or <math>(k,h)</math>, is congruent to the angle <math>(h',k')</math> and at the same time all interior points of the angle <math>(h',k')</math> lie upon the given side of <math>a'</math>. We express this relation by means of the notation
 +
<cmath>\angle (h,k) \cong \angle (h',k')</cmath>
 +
Every angle is congruent to itself; that is,
 +
<cmath>\angle (h,k) \cong \angle (h,k)</cmath>
 +
or
 +
<cmath>\angle (h,k) \cong \angle (k,h)</cmath>
 +
 
 +
*If the angle <math>(h,k)</math> is congruent to the angle <math>(h',k')</math> and to the angle <math>(h'',k'')</math>, then the angle <math>(h',k')</math> is congruent to the angle <math>(h'',k'')</math>; that is to say, if <math>\angle (h, k) \cong \angle (h', k')</math> and
 +
<math>\angle (h, k) \equiv \angle (h'',k'')</math>, then <math>\angle (h',k') \cong \angle (h'',k'')</math>.
 +
 
 +
*If, in the two triangles <math>ABC</math> and <math>A'B'C'</math> the congruences
 +
<cmath>\overline{AB}\cong\overline{A'B'}, \: \overline{AC}\cong\overline{A'C'}, \: \angle BAC\cong\angle B'A'C'</cmath>
 +
hold, then the congruences
 +
<cmath>\angle ABC\cong\angle A'B'C' \:\mbox{and}\; \angle ACB\cong\angle A'C'B'</cmath>
 +
also hold.
 +
 
 +
[http://www.gutenberg.org/files/17384/17384-pdf.pdf Source: gutenberg.org]
 +
 
 +
==[[Triangle]] Congruence==
 +
=== SSS Congruence ===
  
−
<center>[[Image:reflection.PNG]]</center>
+
If the three sides of one triangle are congruent to the corresponding sides of of another triangle, then congruence between the two triangles is established.
  
 +
We start with <math>\triangle ABC</math> and <math>\triangle XYZ</math> shown in the diagram below where <math>\overline{AB}\cong\overline{XY}</math>, <math>\overline{BC}\cong\overline{YZ}</math>, and <math>\overline{AC}\cong\overline{XZ}</math> supposing that <math>\triangle ABC\cong\triangle XYZ</math>.
 +
<asy>
 +
size(500);
 +
pair A,B,C;
 +
A=(0,0); B=(3,2); C=(5,0);
 +
draw(A--B--C--cycle);
 +
label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE);
 +
add(pathticks(A--B)); add(pathticks(B--C,2)); add(pathticks(A--C,3));
 +
pair X,Y,Z;
 +
X=(10,0); Y=(13,2); Z=(15,0);
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draw(X--Y--Z--cycle);
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label("$X$",X,SW); label("$Y$",Y,N); label("$Z$",Z,SE);
 +
add(pathticks(X--Y)); add(pathticks(Y--Z,2)); add(pathticks(X--Z,3));
 +
</asy>
 +
We construct point <math>B'</math> on the opposite side of <math>\overleftrightarrow{AC}</math> from <math>B</math> such that <math>\triangle AB'C\cong\triangle XYZ</math>. Since <math>B'</math> is on the opposite side of <math>\overleftrightarrow{AC}</math> from <math>B</math>, segment <math>\overline{BB'}</math> must intersect the line <math>\overleftrightarrow{AC}</math> at some point <math>D</math>. Updating our diagram gives us
 +
<asy>
 +
size(500);
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pair A,B,C,D,E;
 +
A=(0,0); B=(3,2); C=(5,0); D=(3,-2); E=(3,0);
 +
draw(A--B--C--cycle); draw(A--D--C--cycle); draw(anglemark(D,A,C)); draw(B--D);
 +
label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label ("$B'$",D,S); label("$D$",E,NE);
 +
add(pathticks(A--B)); add(pathticks(B--C,2)); add(pathticks(A--C,3)); add(pathticks(A--D));
 +
pair X,Y,Z;
 +
X=(10,0); Y=(13,2); Z=(15,0);
 +
draw(X--Y--Z--cycle); draw(anglemark(Z,X,Y));
 +
label("$X$",X,SW); label("$Y$",Y,N); label("$Z$",Z,SE);
 +
add(pathticks(X--Y)); add(pathticks(Y--Z,2)); add(pathticks(X--Z,3));
 +
</asy>
 +
We now have several cases depending on the location of point <math>D</math> relative to <math>A</math> and <math>C</math>.
 +
 +
'''<math>D</math> is strictly between <math>A</math> and <math>C</math>, as shown in the diagram above.'''
 +
Therefore, <math>D</math> must be in the interiors of both <math>\angle ABC</math> and <math>\angle AB'C</math>, but since <math>\triangle ABC\cong\triangle XYZ</math> and <math>\triangle AB'C\cong\triangle XYZ</math>, we have <math>\overline{AB}\cong\overline{XY}\cong\overline{AB'}</math> and <math>\overline{BC}\cong\overline{YZ}\cong\overline{B'C}</math>. From this, we find that <math>\triangle BAB'</math> and <math>\triangle BCB'</math> are isosceles, and therefore <math>\angle ABB'\cong\angle AB'B</math> and <math>\angle CBB'\cong\angle CB'B</math>, so <math>\angle ABC=\angle ABB'+\angle CBB'=\angle AB'B+\angle CB'B=\angle AB'C</math>. We have <math>\triangle ABC\cong\triangle AB'C</math> by SAS congruence.
 +
 +
<!-- stub please edit -->
 +
'''<math>A</math> is strictly between <math>C</math> and <math>D</math>.'''
 +
Without loss of generality, we can derive <math>\triangle ABC\cong\triangle AB'C</math> from the fact that <math>C</math> is strictly between <math>A</math> and <math>D</math>.
 +
 +
{{stub}}
 +
=== SAS Congruence ===
 +
 +
The SAS Congruence theorem is derived from the sixth axiom of congruence. In short, the sixth axiom states that when given two triangles, if two corresponding side congruences hold and the angle between the two sides is equal on both triangles, then the other two angles of the triangle are equal.
 +
 +
{{stub}}
 +
 +
=== ASA Congruence ===
 +
 +
{{stub}}
 +
 +
=== AAS Congruence ===
 +
 +
Since the third angle of a triangle is always the difference of <math>180^\circ</math> and the other two angles, two triangles with two pairs of congruent angles would give another pair of congruent angles. When we have a pair of congruent sides and two pairs of congruent angles adjacent to the side, it is ASA congruence. However, if we're given a pair of congruent angles opposite from the side that is congruent and a pair of congruent angles adjacent to that side, everything is congruent. Therefore, AAS congruence can be proven with ASA congruence.
 +
 +
{{stub}}
 +
 +
=== HL Congruence ===
 +
 +
If both the hypotenuse and leg of one right triangle are congruent to that of another, the two triangles are congruent.
 +
 +
Consider right <math>\triangle ABC</math> and right <math>\triangle XYZ</math> shown in the diagram below. We are given that <math>AB=XY</math> and <math>AC=XZ</math>.
 +
<asy>
 +
size(500);
 +
pair A,B,C;
 +
A=(0,0); B=(9/5,12/5); C=(5,0);
 +
draw(A--B--C--cycle);
 +
label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE);
 +
draw(rightanglemark(A,B,C));
 +
pair X,Y,Z;
 +
X=(10,0); Y=(59/5,12/5); Z=(15,0);
 +
draw(X--Y--Z--cycle);
 +
label("$X$",X,SW); label("$Y$",Y,N); label("$Z$",Z,SE);
 +
draw(rightanglemark(X,Y,Z));
 +
</asy>
 +
Since <math>\angle B=90^\circ</math>, the Pythagorean theorem gives us <math>AB^2+BC^2=AC^2</math>. Similarly, using the Pythagorean theorem on <math>\triangle XYZ</math> gives us <math>XY^2+YZ^2=XZ^2</math>. However, since <math>AB=XY</math> and <math>AC=XZ</math>, substitution in the second equation gives <math>AB^2+YZ^2=AC^2</math>. Subtracting this from our first equation plus a little manipiulation gives us <math>AC^2=YZ^2</math>. Therefore, since all lengths are positive, taking the square root of both sides gives <math>AC=YZ</math>, so, by the SSS congruence theorem, we have <math>\triangle ABC\cong\triangle XYZ</math>.
 +
 +
Since the congruent angle given is not between the two equivalent sides, this may be seen as SSA congruence, which is not necessarily correct. However, this form of SSA congruence holds true for right triangles.
 +
 +
=== LL Congruence ===
 +
 +
LL Congruence is the basically the same as SAS congruence since we are given a leg, a right angle, and the other leg.
  
 
== See also ==
 
== See also ==
Line 16: Line 130:
  
 
[[Category:Definition]]
 
[[Category:Definition]]
 +
 +
[[Category:Geometry]]

Latest revision as of 15:56, 17 September 2024

Congruency is a property of multiple geometric figures.

Intuitive Definition

Two geometric figures are congruent if one of them can be turned and/or flipped and placed exactly on top of the other, with all parts lining up perfectly with no parts on either figure left over. In plain language, two objects are congruent if they have the same size and shape.

A collection of isometries.

Technical Definition

Two geometric objects are congruent if one can be transformed into the other by an isometry, such as a translation, rotation, reflection or some combination thereof.

Axioms

  • If $A$, $B$ are two points on a straight line $a$, and if $A'$ is a point upon the same or another straight line $a'$, then, upon a given side of $A'$ on the straight line $a'$, we can always find one and only one point $B'$ so that the segment $AB$ (or $BA$) is congruent to the segment $A'B'$. We indicate this relation by writing

\[\overline{AB}\cong\overline{A'B'}.\] Every segment is congruent to itself; that is, we always have \[\overline{AB}\cong\overline{AB}.\]

  • If a segment $\overline{AB}$ is congruent to the segment $\overline{A'B'}$ and also to the segment $\overline{A''B''}$, then the segment $\overline{A'B'}$ is congruent to the segment $\overline{A''B''}$; that is, if $\overline{AB}\cong\overline{A'B}$ and $\overline{AB}\cong\overline{A''B''}$, then $\overline{A'B'}\cong\overline{A''B''}$.
  • Let $\overline{AB}$ and $\overline{BC}$ be two segments of a straight line $a$ which have no points in common aside from the point $B$, and, furthermore, let $\overline{A'B'}$ and $\overline{B'C'}$ be two segments of the same or of another straight line $a'$ having, likewise, no point other than $B'$ in common. Then, if $\overline{AB}\cong\overline{A'B'}$ and $\overline{BC}\cong\overline{B'C'}$, we have $\overline{AC}\cong\overline{A'C'}$.
  • Let an angle $(h,k)$ be given in the plane $\alpha$ and let a straight line $a'$ be given in a plane $\alpha'$. Suppose also that, in the plane $\alpha$, a definite side of the straight line $a'$ be assigned. Denote by $h'$ a half-ray of the straight line $a'$ emanating from a point $O'$ of this line. Then in the plane $\alpha'$ there is one and only one half-ray $k'$ such that the angle $(h,k)$, or $(k,h)$, is congruent to the angle $(h',k')$ and at the same time all interior points of the angle $(h',k')$ lie upon the given side of $a'$. We express this relation by means of the notation

\[\angle (h,k) \cong \angle (h',k')\] Every angle is congruent to itself; that is, \[\angle (h,k) \cong \angle (h,k)\] or \[\angle (h,k) \cong \angle (k,h)\]

  • If the angle $(h,k)$ is congruent to the angle $(h',k')$ and to the angle $(h'',k'')$, then the angle $(h',k')$ is congruent to the angle $(h'',k'')$; that is to say, if $\angle (h, k) \cong \angle (h', k')$ and

$\angle (h, k) \equiv \angle (h'',k'')$, then $\angle (h',k') \cong \angle (h'',k'')$.

  • If, in the two triangles $ABC$ and $A'B'C'$ the congruences

\[\overline{AB}\cong\overline{A'B'}, \: \overline{AC}\cong\overline{A'C'}, \: \angle BAC\cong\angle B'A'C'\] hold, then the congruences \[\angle ABC\cong\angle A'B'C' \:\mbox{and}\; \angle ACB\cong\angle A'C'B'\] also hold.

Source: gutenberg.org

Triangle Congruence

SSS Congruence

If the three sides of one triangle are congruent to the corresponding sides of of another triangle, then congruence between the two triangles is established.

We start with $\triangle ABC$ and $\triangle XYZ$ shown in the diagram below where $\overline{AB}\cong\overline{XY}$, $\overline{BC}\cong\overline{YZ}$, and $\overline{AC}\cong\overline{XZ}$ supposing that $\triangle ABC\cong\triangle XYZ$. [asy] size(500); pair A,B,C; A=(0,0); B=(3,2); C=(5,0); draw(A--B--C--cycle); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); add(pathticks(A--B)); add(pathticks(B--C,2)); add(pathticks(A--C,3)); pair X,Y,Z; X=(10,0); Y=(13,2); Z=(15,0); draw(X--Y--Z--cycle); label("$X$",X,SW); label("$Y$",Y,N); label("$Z$",Z,SE); add(pathticks(X--Y)); add(pathticks(Y--Z,2)); add(pathticks(X--Z,3)); [/asy] We construct point $B'$ on the opposite side of $\overleftrightarrow{AC}$ from $B$ such that $\triangle AB'C\cong\triangle XYZ$. Since $B'$ is on the opposite side of $\overleftrightarrow{AC}$ from $B$, segment $\overline{BB'}$ must intersect the line $\overleftrightarrow{AC}$ at some point $D$. Updating our diagram gives us [asy] size(500); pair A,B,C,D,E; A=(0,0); B=(3,2); C=(5,0); D=(3,-2); E=(3,0); draw(A--B--C--cycle); draw(A--D--C--cycle); draw(anglemark(D,A,C)); draw(B--D); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label ("$B'$",D,S); label("$D$",E,NE); add(pathticks(A--B)); add(pathticks(B--C,2)); add(pathticks(A--C,3)); add(pathticks(A--D)); pair X,Y,Z; X=(10,0); Y=(13,2); Z=(15,0); draw(X--Y--Z--cycle); draw(anglemark(Z,X,Y)); label("$X$",X,SW); label("$Y$",Y,N); label("$Z$",Z,SE); add(pathticks(X--Y)); add(pathticks(Y--Z,2)); add(pathticks(X--Z,3)); [/asy] We now have several cases depending on the location of point $D$ relative to $A$ and $C$.

$D$ is strictly between $A$ and $C$, as shown in the diagram above. Therefore, $D$ must be in the interiors of both $\angle ABC$ and $\angle AB'C$, but since $\triangle ABC\cong\triangle XYZ$ and $\triangle AB'C\cong\triangle XYZ$, we have $\overline{AB}\cong\overline{XY}\cong\overline{AB'}$ and $\overline{BC}\cong\overline{YZ}\cong\overline{B'C}$. From this, we find that $\triangle BAB'$ and $\triangle BCB'$ are isosceles, and therefore $\angle ABB'\cong\angle AB'B$ and $\angle CBB'\cong\angle CB'B$, so $\angle ABC=\angle ABB'+\angle CBB'=\angle AB'B+\angle CB'B=\angle AB'C$. We have $\triangle ABC\cong\triangle AB'C$ by SAS congruence.

$A$ is strictly between $C$ and $D$. Without loss of generality, we can derive $\triangle ABC\cong\triangle AB'C$ from the fact that $C$ is strictly between $A$ and $D$.

This article is a stub. Help us out by expanding it.

SAS Congruence

The SAS Congruence theorem is derived from the sixth axiom of congruence. In short, the sixth axiom states that when given two triangles, if two corresponding side congruences hold and the angle between the two sides is equal on both triangles, then the other two angles of the triangle are equal.

This article is a stub. Help us out by expanding it.

ASA Congruence

This article is a stub. Help us out by expanding it.

AAS Congruence

Since the third angle of a triangle is always the difference of $180^\circ$ and the other two angles, two triangles with two pairs of congruent angles would give another pair of congruent angles. When we have a pair of congruent sides and two pairs of congruent angles adjacent to the side, it is ASA congruence. However, if we're given a pair of congruent angles opposite from the side that is congruent and a pair of congruent angles adjacent to that side, everything is congruent. Therefore, AAS congruence can be proven with ASA congruence.

This article is a stub. Help us out by expanding it.

HL Congruence

If both the hypotenuse and leg of one right triangle are congruent to that of another, the two triangles are congruent.

Consider right $\triangle ABC$ and right $\triangle XYZ$ shown in the diagram below. We are given that $AB=XY$ and $AC=XZ$. [asy] size(500); pair A,B,C; A=(0,0); B=(9/5,12/5); C=(5,0); draw(A--B--C--cycle); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); draw(rightanglemark(A,B,C)); pair X,Y,Z; X=(10,0); Y=(59/5,12/5); Z=(15,0); draw(X--Y--Z--cycle); label("$X$",X,SW); label("$Y$",Y,N); label("$Z$",Z,SE); draw(rightanglemark(X,Y,Z)); [/asy] Since $\angle B=90^\circ$, the Pythagorean theorem gives us $AB^2+BC^2=AC^2$. Similarly, using the Pythagorean theorem on $\triangle XYZ$ gives us $XY^2+YZ^2=XZ^2$. However, since $AB=XY$ and $AC=XZ$, substitution in the second equation gives $AB^2+YZ^2=AC^2$. Subtracting this from our first equation plus a little manipiulation gives us $AC^2=YZ^2$. Therefore, since all lengths are positive, taking the square root of both sides gives $AC=YZ$, so, by the SSS congruence theorem, we have $\triangle ABC\cong\triangle XYZ$.

Since the congruent angle given is not between the two equivalent sides, this may be seen as SSA congruence, which is not necessarily correct. However, this form of SSA congruence holds true for right triangles.

LL Congruence

LL Congruence is the basically the same as SAS congruence since we are given a leg, a right angle, and the other leg.

See also

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