Let’s prove that the measurement of the angle made by the bisectors of two adjacent angles of quadrilateral is equal to the half of the sum of measurement of the other two angles.

Let us first draw a diagram for the above question.

ABCD is the required quadrilateral.

OA is the angle bisector for ∠ DAB and OB is the angle bisector for ∠ ABC.

When we observe the above figure we know that,

∠ BAO = ∠ OAB = � ∠ BAD [as OA is the angle bisector of ∠ DAB].. (i)

∠ OBC = ∠ ABO = � ∠ ABC [as OB is the angle bisector of ∠ ABC] .. (ii)

To Prove: ∠ AOB = (∠ BCD + ∠ CDA) / 2

Proof:

For a quadrilateral we know that the sum of all interior angles is 360°.

Even we can verify it using the formula.

Sum of all interior angles = (2n – 4) × 90

= (2 × 4 – 4) × 90

= (8 – 4) × 90

= 4 × 90

= 360°

So ∠ DAB + ∠ ABC + ∠ BCD + ∠ CDA = 360° ………. (iii)

For a triangle the sum of all interior angles is 180°

So for Δ AOB

∠ ABO + ∠ OAB + ∠ AOB = 180°

From (i) we have ∠ OAB = � ∠ BAD and from (ii) we have ∠ ABO = � ∠ ABC.

So let us substitute these values in above equation.

� ∠ BAD + � ∠ ABC + ∠ AOB = 180°

∠ BAD + ∠ ABC + 2 × ∠ AOB = 2 × 180° [Removing the fraction part by multiplying throughout by 2]

∠ BAD + ∠ ABC + 2 × ∠ AOB = 360° …………. (iv)

Now for equation (iii) and (iv) the right hand side is same. Hence the left hand side has to be equal.

So let us equate the two left hand sides parts.

∠ BAD + ∠ ABC + 2 × ∠ AOB = ∠ DAB + ∠ ABC + ∠ BCD + ∠ CDA

∠ BAD and ∠ ABC gets eliminated from both the sides.

2 × ∠ AOB = ∠ BCD + ∠ CDA

∠ AOB = � (∠ BCD + ∠ CDA)

Hence proved that that the measurement of the angle made by the bisectors of two adjacent angles of quadrilateral is equal to the half of the sum of measurement of the other two angles.