From the figure, we know that,
AB and CD intersect each other at point O.
Let the two pairs of vertically opposite angles be,
1st pair - ∠AOC and ∠BOD
2nd pair - ∠AOD and ∠BOC
To prove:
Vertically opposite angles are equal,
i.e., ∠AOC = ∠BOD, and ∠AOD = ∠BOC
From the figure,
The ray AO stands on the line CD.
We know that,
If a ray lies on a line then the sum of the adjacent angles is equal to 180°.
⇒ ∠AOC + ∠AOD = 180° (By linear pair axiom) … (i)
Similarly, the ray DO lies on the line AOB.
⇒ ∠AOD + ∠BOD = 180° (By linear pair axiom) … (ii)
From equations (i) and (ii),
We have,
∠AOC + ∠AOD = ∠AOD + ∠BOD
⇒ ∠AOC = ∠BOD - - - - (iii)
Similarly, the ray BO lies on the line COD.
⇒ ∠DOB + ∠COB = 180° (By linear pair axiom) - - - - (iv)
Also, the ray CO lies on the line AOB.
⇒ ∠COB + ∠AOC = 180° (By linear pair axiom) - - - - (v)
From equations (iv) and (v),
We have,
∠DOB + ∠COB = ∠COB + ∠AOC
⇒ ∠DOB = ∠AOC - - - - (vi)
Thus, from equation (iii) and equation (vi),
We have,
∠AOC = ∠BOD, and ∠DOB = ∠AOC
Therefore, we get, vertically opposite angles are equal.
Hence Proved.
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Question 1 If two lines intersect, prove that the vertically opposite angles are equal.
7
Answer
Note: We can also prove that the other pair of vertically opposite angles $\angle COB$ and $\angle DOA$ equal in the similar way as shown below:
Let us consider the angles on the line CD. So, we get $\angle COB+\angle BOD={{180}^{\circ }}$.$\Rightarrow \angle BOD={{180}^{\circ }}-\angle COB$ ---(3).Now, let us consider the angles on the line AB. So, we get $\angle BOD+\angle DOA={{180}^{\circ }}$ ---(4).Let us substitute equation (3) in equation (4).So, we get ${{180}^{\circ }}-\angle COB+\angle DOA={{180}^{\circ }}$.$\Rightarrow \angle DOA={{180}^{\circ }}-{{180}^{\circ }}+\angle COB$.$\Rightarrow \angle DOA=\angle COB$.We use this result to get the required answers.