Vertical angles can be defined as the pairs of non-adjacent angles in opposite positions formed by the intersection of two straight lines are known as vertical angles.
Vertical Angles
In the figure, $\angle$1 and $\angle$3 are a pair of vertical angles. Another pair of vertical angles is $\angle$2 and $\angle$4.

If two angles are vertical angles, then they have equal measures.

Construction: Let $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ be two intersecting lines in a plane, intersecting at the point P, as shown in the figure. Here, the angles $\angle$APD and $\angle$BPC are vertical angles.
Vertical Angles Theorem

To Prove:
$$\angle APD \cong \angle BPC$$
S. No.StatementReason
1∠APD+∠APC = 180oAngles on a straight line (Supplementary angles)
2∠BPC+∠APC = 180oAngles on a straight line
3∠APD+∠APC=∠BPC+∠APCFrom (1) & (2)
4∠APC≅∠BPC(3), Subtracted ∠APC from both sides

Hence, we proved the theorem.
The following are some of the real life example where we can identify vertical angles.

Example 1: At the crossing of two straight roads, we can always find two pairs of vertical angles. One of those two pairs is shown in the diagram.

Vertical Angles in Real Life

Example 2: The office is divided into cabins. The non-adjacent opposite angles at the corners of the cabins are vertical angles.

Solved Example

Question: Find the values of x and y in the following figure.

How to Solve Vertical Angles
From the figure, it is clear that (2x)o and 70o are vertical angles. So, by the Vertical Angles Theorem, we have
          2x = 70
        $\rightarrow$x = 35

From the figure, it is also clear that the angles (2x+y)o and 70o are angles on a straight line. So, they are supplementary angles.  Thus, we have
        (2x + y)o + 70o = 180o
        $\rightarrow$ 2x + y = 180 - 70
        $\rightarrow$ 2(35) + y = 110
        $\rightarrow$ 70 + y = 110
        $\rightarrow$ y = 110 - 70
        $\rightarrow$ y = 40
Thus, we have
        x = 35 and y = 40.

The following are two examples on proving the given statements using vertical angles and the Vertical Angles Theorem.

Solved Examples

Question 1: If A, B, and C are three non-collinear points, then there exists a point D that does not lie on $\overleftrightarrow{AB}$ such that the angle measure of one of the interior angles in $
\Delta$ABD is less than or equal to $\frac{1}{2}$  $\angle$CAB.
Hypothesis:  Let A, B, and C be three non-collinear points.

Construction: Let E be the mid-point of $\overline{BC}$. Let D be a point on  $\overrightarrow{AE}$ such that E is the mid-point of $\overline{AD}$.  That is, AE = ED.
Clearly, $\angle$AEC and $\angle$DEB are vertical angles.

Vertical Angles Examples
To Prove:
Either $\angle$BAD $\leq$ $\frac{1}{2}$ $\angle$BAC or $\angle$ADB $\leq$ $\frac{1}{2}$ $\angle$BAC
S. No.Statement Reason
1AE=DEE is the mid-point of AD
2CE=BEE is the mid-point of BC
3∠AEC≅∠DEBBy Vertical Angles Theorem
4ΔAEC≅ΔDEB(1), (2), (3) and By SAS Property of congruency
5∠CAE≅∠ADBCorresponding angles in congruent triangles
6∠BAE≅∠BADSame angle
7∠BAC=∠BAE+∠CAESum of angles
8∠BAC=∠BAD+∠ADB(5), (6) and (7)
9∠BAD < $\frac{1}{2}$ ∠ BAC or ∠ADB ≤ $\frac{1}{2}$(8)


Question 2: Every triangle is equivalent by dissection to a rectangle.
Hypothesis: Let $\Delta$ABC be a triangle.

Construction:  Suppose that $\angle$A and $\angle$B are acute angles in the triangle $\Delta$ABC (Re-label the triangle if needed).  Let M be the mid-point of $\overline{AC}$ and N be the mid-point of $\overline{BC}$. Draw the line $\overleftrightarrow{MN}$ through M and N.  Let D and E be the foot of the perpendiculars of A and B on the line $\overleftrightarrow{MN}$, respectively; and let F be the foot of the perpendicular of C on the line $\overleftrightarrow{MN}$.
From the figure, it is clear that $\angle$DMA and $\angle$FMC area a pair of vertical angles. And, $\angle$CNF and $\angle$BNE are another pair of vertical angles.

To Prove:
        Area of $\square$ABED = Area of $\Delta$ABC.


S. No.StatementReason
1∠DMA≅∠FMCBy Vertical Angle Theorem
2∠MDA≅∠MFCRight angles
3AM=CMM is the mid-point of $\overline{AC}$
4ΔADM≅ΔCFMBy AAS property of congruency
5Area of ΔADM = Area of ΔCFMAreas of congruence triangles
6∠CNF≅∠BNEBy Vertical Angle Theorem
7∠CFN≅∠BENRight angles
8CN=BNN is the mid-point of $\overline{BC}$
9ΔCFN≅ΔBENBy AAS property of congruency
10Area of ΔCFN = Area of ΔBENAreas of congruence triangles
11Area of ▭ABED = Area of ΔBEN+Area of ▭ABNM+Area of ΔADMArea of whole is equal to sum of the areas of parts
12Area of ▭ABED = Area of ΔCFN+Area of ▭ABNM+Area of ΔCFMUsing (5) and (10) in (11)
13Area of ΔABC= Area of ΔCFN+Area of ▭ABNM+Area of ΔCFMArea of whole is equal to sum of the areas of parts
14Area of ▭ABED = Area of ΔABCBy (12) and (13)

Hence, we proved the statement.