Question
Question: \[{{C}_{1}}\]and \[{{C}_{2}}\] are two circles with center \[{{O}_{1}}\] and \[{{O}_{2}}\] and inter...
C1and C2 are two circles with center O1 and O2 and intersect each other at points Aand B. If O1O2 intersectAB atM then show that ΔO1AO2≅ΔO1BO2
Solution
Hint: We will try to show the two triangles ΔO1AB and ΔO2AB congruent using ‘SSS’ type of triangle congruence.
Given that two circles C1and C2 with center O1and O2 intersect each other at points Aand B.
Also, O1O2 intersects AB at M.
Then, we have to show that ΔO1AO2≅ΔO1BO2
Let us assume that the radius of the circle C1be r and the radius of the circle C2 be s.
Proof:
In ΔO1AO2 and ΔO1BO2, we have
O1A=O1B.....(i)
Both are radii of the same circle C1.
⇒O2A=O2B.....(ii)
Both are radii of the same circle C2.
Also, O1O2=O2O1....(iii)
Common sides of ΔO1AO2 and ΔO1BO2
So, from equation (i),(ii)and(iii), we get both triangles ΔO1AO2and ΔO1BO2 are congruent with each other by ‘SSS’ type if triangle congruence.
Or,ΔO1AO2≅ΔO1BO2 by SSS type of triangle congruence.
(Here, ‘SSS’ type means side – side – side type of triangle congruence)
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 ΔABC and ΔDEF, we can say that
AB=DF....(a)
AC=DE....(b)
BC=EF....(c)
So, from equation (a),(b) and (c), we have ΔABC≅ΔDEF by SSS – type triangle congruence.
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 theorem, we prove the congruence of the required triangles to prove what is given.