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.

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.

## Vertical Angles Theorem

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.

To Prove:
$$\angle APD \cong \angle BPC$$
Proof:
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.

## Vertical Angles in Real Life

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.

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

## How to Solve Vertical Angles?

### Solved Example

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

Solution:

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.

## Vertical Angles Examples

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.
Solution:

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.

To Prove:
Either $\angle$BAD $\leq$ $\frac{1}{2}$ $\angle$BAC or $\angle$ADB $\leq$ $\frac{1}{2}$ $\angle$BAC
Proof:
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
7∠BAC=∠BAE+∠CAESum of angles
9∠BAD < $\frac{1}{2}$ ∠ BAC or ∠ADB ≤ $\frac{1}{2}$(8)

Question 2: Every triangle is equivalent by dissection to a rectangle.
Solution:

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.

Proof:

S. No.StatementReason
1∠DMA≅∠FMCBy Vertical Angle Theorem
2∠MDA≅∠MFCRight angles
3AM=CMM is the mid-point of $\overline{AC}$
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.