\[{{C}_{1}}\]and \[{{C}_{2}}\] are two circles with center \[{{O}_{1}}\] and \[{{O}_{2}}\] and intersect each other at points \[A\] and \[B\]. If \[{{O}_{1}}{{O}_{2}}\] intersect \[AB\] at \[M\] then show that \[M\] is the midpoint of \[AB\].

Last updated date: 29th Mar 2023
•
Total views: 309.6k
•
Views today: 4.90k
Answer
309.6k+ views
Hint: We will try to show \[\Delta {{O}_{1}}AB\]\[\cong \]\[\Delta {{O}_{2}}AB\] using ‘SSS’ type of triangle congruency and then we will show \[\Delta {{O}_{1}}AM\cong \Delta {{O}_{1}}BM\] using ‘SAS’ type of triangle congruency and finally we will show \[M\] is the midpoint of \[AB\].
Given that two circles \[{{C}_{1}}\]and \[{{C}_{2}}\] with the center \[{{O}_{1}}\] and \[{{O}_{2}}\] intersect each other at points \[A\] and \[B\].
Also, \[{{O}_{1}}{{O}_{2}}\] intersects \[AB\] at \[M\].
Then, we have to show that \[M\] is the midpoint of \[AB\].
Let us assume that the radius of the circle \[{{C}_{1}}\] be \[r\] and the radius of the circle \[{{C}_{2}}\] be \[s\].
Proof:
In \[\Delta {{O}_{1}}A{{O}_{2}}\] and \[\Delta {{O}_{1}}B{{O}_{2}}\], we have
\[{{O}_{1}}A={{O}_{1}}B.....\left( i \right)\]
Both are radii of the same circle \[{{C}_{1}}\].
\[\Rightarrow {{O}_{2}}A={{O}_{2}}B.....\left( ii \right)\]
Both are radii of the same circle \[{{C}_{2}}\].
Also, \[{{O}_{1}}{{O}_{2}}={{O}_{2}}{{O}_{1}}....\left( iii \right)\]
Common sides of both the triangles \[\Delta {{O}_{1}}A{{O}_{2}}\]and \[\Delta {{O}_{1}}B{{O}_{2}}\]
So, from the equation \[\left( i \right),\left( ii \right)\]and\[\left( iii \right)\], we get both triangles \[\Delta {{O}_{1}}A{{O}_{2}}\] and \[\Delta {{O}_{1}}B{{O}_{2}}\] are congruent with each other by ‘SSS’ type of triangle congruency.
Or, \[\Delta {{O}_{1}}A{{O}_{2}}\cong \Delta {{O}_{1}}B{{O}_{2}}\] by SSS type of triangle congruency.
(Here, ‘SSS’ type means side – side – side type of triangle congruency)
SSS – Theorem
Side-side - side postulate (SSS) states that two triangles are congruent if three sides of one triangle are congruent to the corresponding sides of the other triangle.
Here, from \[\Delta ABC\] and \[\Delta DEF\], we can say that
\[AB=DF....\left( a \right)\]
\[AC=DE....\left( b \right)\]
\[BC=EF....\left( c \right)\]
So, from the equation \[\left( a \right),\left( b \right)\] and \[\left( c \right)\], we have \[\Delta ABC\cong \Delta DEF\] by SSS – type triangle congruency which clearly shows that
\[\Rightarrow M\angle {{O}_{2}}{{O}_{1}}A=M\angle {{O}_{2}}{{O}_{1}}B....\left( iv \right)\] by CPCT
(Here, CPCT is corresponding parts of the congruent triangle)
Also, we have,
\[\Rightarrow M\angle M{{O}_{1}}A=M\angle M{{O}_{1}}B....\left( v \right)\]
From the equation \[\left( iv \right)\], both are the same angle.
\[\Rightarrow {{O}_{1}}A={{O}_{1}}B.....\left( vi \right)\] by CPCT
Now, in \[\Delta AM{{O}_{1}}\]and \[\Delta BM{{O}_{1}}\], we have
\[\Rightarrow {{O}_{1}}A={{O}_{1}}B....\left( vii \right)\]
(Both are radii of the same circle)
\[\Rightarrow m\angle M{{O}_{1}}A=M\angle M{{O}_{1}}B.....\left( viii \right)\]
From corresponding parts of the congruent triangle.
\[\Rightarrow {{O}_{1}}M=M{{O}_{1}}.....\left( ix \right)\]
The common side of both the triangles \[\Delta AM{{O}_{1}}\]and \[\Delta BM{{O}_{1}}\].
So, from the equation \[\left( vii \right),\left( viii \right)\]and \[\left( ix \right)\], we get both triangles \[\Delta AM{{O}_{1}}\]and \[\Delta BM{{O}_{1}}\] are congruent with each other by the SAS test of triangle congruency.
Or, \[\Delta AM{{O}_{1}}\cong BM{{O}_{1}}\] by ’SAS’ test of triangle congruency.
(Here, ‘SAS’ means Side – Angle – Side type of triangle congruency)
SAS – theorem
If any two sides and the angle included between one triangle are equivalent to the corresponding two sides and the angle between the sides of the second triangle, then the two triangles are said to be congruent by SAS rule.
Here in the diagram,
Side \[AB=DE\]
Also side \[BC=EF\]
And \[\angle ABC=\angle DEF\]
Thus, \[\Delta ABC\cong \Delta DEF\] by SAS type of triangle congruency.
Also, from this, we have\[AM=BM\] by the corresponding part of congruent triangles \[\Delta AM{{O}_{1}}\cong \Delta BM{{O}_{{}}}\] which means that \[M\] is the midpoint of \[AB\].
Hence proved.
Note: Visualize the geometry first before attempting the question. Make a clear diagram of the required question which reduces the probability of error in your solution using the SSS and SAS theorems to prove triangles are congruent, we prove the required statement.
Given that two circles \[{{C}_{1}}\]and \[{{C}_{2}}\] with the center \[{{O}_{1}}\] and \[{{O}_{2}}\] intersect each other at points \[A\] and \[B\].
Also, \[{{O}_{1}}{{O}_{2}}\] intersects \[AB\] at \[M\].
Then, we have to show that \[M\] is the midpoint of \[AB\].

Let us assume that the radius of the circle \[{{C}_{1}}\] be \[r\] and the radius of the circle \[{{C}_{2}}\] be \[s\].
Proof:
In \[\Delta {{O}_{1}}A{{O}_{2}}\] and \[\Delta {{O}_{1}}B{{O}_{2}}\], we have
\[{{O}_{1}}A={{O}_{1}}B.....\left( i \right)\]
Both are radii of the same circle \[{{C}_{1}}\].
\[\Rightarrow {{O}_{2}}A={{O}_{2}}B.....\left( ii \right)\]
Both are radii of the same circle \[{{C}_{2}}\].
Also, \[{{O}_{1}}{{O}_{2}}={{O}_{2}}{{O}_{1}}....\left( iii \right)\]
Common sides of both the triangles \[\Delta {{O}_{1}}A{{O}_{2}}\]and \[\Delta {{O}_{1}}B{{O}_{2}}\]
So, from the equation \[\left( i \right),\left( ii \right)\]and\[\left( iii \right)\], we get both triangles \[\Delta {{O}_{1}}A{{O}_{2}}\] and \[\Delta {{O}_{1}}B{{O}_{2}}\] are congruent with each other by ‘SSS’ type of triangle congruency.
Or, \[\Delta {{O}_{1}}A{{O}_{2}}\cong \Delta {{O}_{1}}B{{O}_{2}}\] by SSS type of triangle congruency.
(Here, ‘SSS’ type means side – side – side type of triangle congruency)
SSS – Theorem

Side-side - side postulate (SSS) states that two triangles are congruent if three sides of one triangle are congruent to the corresponding sides of the other triangle.
Here, from \[\Delta ABC\] and \[\Delta DEF\], we can say that
\[AB=DF....\left( a \right)\]
\[AC=DE....\left( b \right)\]
\[BC=EF....\left( c \right)\]
So, from the equation \[\left( a \right),\left( b \right)\] and \[\left( c \right)\], we have \[\Delta ABC\cong \Delta DEF\] by SSS – type triangle congruency which clearly shows that
\[\Rightarrow M\angle {{O}_{2}}{{O}_{1}}A=M\angle {{O}_{2}}{{O}_{1}}B....\left( iv \right)\] by CPCT
(Here, CPCT is corresponding parts of the congruent triangle)
Also, we have,
\[\Rightarrow M\angle M{{O}_{1}}A=M\angle M{{O}_{1}}B....\left( v \right)\]
From the equation \[\left( iv \right)\], both are the same angle.
\[\Rightarrow {{O}_{1}}A={{O}_{1}}B.....\left( vi \right)\] by CPCT
Now, in \[\Delta AM{{O}_{1}}\]and \[\Delta BM{{O}_{1}}\], we have
\[\Rightarrow {{O}_{1}}A={{O}_{1}}B....\left( vii \right)\]
(Both are radii of the same circle)
\[\Rightarrow m\angle M{{O}_{1}}A=M\angle M{{O}_{1}}B.....\left( viii \right)\]
From corresponding parts of the congruent triangle.
\[\Rightarrow {{O}_{1}}M=M{{O}_{1}}.....\left( ix \right)\]
The common side of both the triangles \[\Delta AM{{O}_{1}}\]and \[\Delta BM{{O}_{1}}\].
So, from the equation \[\left( vii \right),\left( viii \right)\]and \[\left( ix \right)\], we get both triangles \[\Delta AM{{O}_{1}}\]and \[\Delta BM{{O}_{1}}\] are congruent with each other by the SAS test of triangle congruency.
Or, \[\Delta AM{{O}_{1}}\cong BM{{O}_{1}}\] by ’SAS’ test of triangle congruency.
(Here, ‘SAS’ means Side – Angle – Side type of triangle congruency)
SAS – theorem

If any two sides and the angle included between one triangle are equivalent to the corresponding two sides and the angle between the sides of the second triangle, then the two triangles are said to be congruent by SAS rule.
Here in the diagram,
Side \[AB=DE\]
Also side \[BC=EF\]
And \[\angle ABC=\angle DEF\]
Thus, \[\Delta ABC\cong \Delta DEF\] by SAS type of triangle congruency.
Also, from this, we have\[AM=BM\] by the corresponding part of congruent triangles \[\Delta AM{{O}_{1}}\cong \Delta BM{{O}_{{}}}\] which means that \[M\] is the midpoint of \[AB\].
Hence proved.
Note: Visualize the geometry first before attempting the question. Make a clear diagram of the required question which reduces the probability of error in your solution using the SSS and SAS theorems to prove triangles are congruent, we prove the required statement.
Recently Updated Pages
If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts
What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?
