Given a tangential quadrilateral $ABCD$, and $O$ is the center of the inscribed circle, prove that $\angle AOB + \angle COD = 180^\circ$.

1. Problem Description

Given a tangential quadrilateral $ABCD$, and $O$ is the center of the inscribed circle, prove that $\angle AOB + \angle COD = 180^\circ$.

2. Solution Steps

Let the points where the inscribed circle touches the sides $AB, BC, CD, DA$ be $P, Q, R, S$ respectively.

Since $O$ is the center of the inscribed circle, $OP \perp AB, OQ \perp BC, OR \perp CD, OS \perp DA$.

Consider the quadrilateral $APOS$. We have $\angle APS = \angle ASO = 90^{\circ}$. The sum of the angles in a quadrilateral is $360^{\circ}$, so

$\angle POS = 360^{\circ} - \angle ASO - \angle APS - \angle A = 360^{\circ} - 90^{\circ} - 90^{\circ} - \angle A = 180^{\circ} - \angle A$.

Similarly, for quadrilateral $BPOQ$, $\angle BOQ = 180^{\circ} - \angle B$.

For quadrilateral $CROQ$, $\angle COR = 180^{\circ} - \angle C$.

For quadrilateral $DROS$, $\angle DOS = 180^{\circ} - \angle D$.

The sum of angles around point $O$ is $360^{\circ}$, so

$\angle AOB + \angle BOC + \angle COD + \angle DOA = 360^{\circ}$.

$\angle AOB = \angle AOP + \angle POB$.

Let $\angle AOB$ be the angle subtended by side $AB$ at the center $O$. Similarly, $\angle BOC, \angle COD, \angle DOA$ are angles subtended by the sides $BC, CD, DA$ at the center $O$ respectively.

Since $ABCD$ is a tangential quadrilateral, $AB + CD = BC + DA$.

Also, we know that the sum of angles in a quadrilateral is $360^{\circ}$, so $\angle A + \angle B + \angle C + \angle D = 360^{\circ}$.

We want to prove that $\angle AOB + \angle COD = 180^{\circ}$.

Consider the sum of angles around point $O$. We have $\angle AOB + \angle BOC + \angle COD + \angle DOA = 360^{\circ}$.

We need to show that $\angle BOC + \angle DOA = 180^{\circ}$.

Since $O$ is the incenter, $AO, BO, CO, DO$ are angle bisectors of $\angle A, \angle B, \angle C, \angle D$ respectively.

Let $\angle OAB = \angle OAS = a$, $\angle OBA = \angle OBP = b$, $\angle OCD = \angle OCR = c$, $\angle ODC = \angle ODR = d$.

Then $\angle A = 2a, \angle B = 2b, \angle C = 2c, \angle D = 2d$.

We know that $2a + 2b + 2c + 2d = 360^{\circ}$, so $a + b + c + d = 180^{\circ}$.

In triangle $AOB$, $\angle AOB = 180^{\circ} - a - b$.

In triangle $COD$, $\angle COD = 180^{\circ} - c - d$.

Thus $\angle AOB + \angle COD = 180^{\circ} - a - b + 180^{\circ} - c - d = 360^{\circ} - (a + b + c + d) = 360^{\circ} - 180^{\circ} = 180^{\circ}$.

3. Final Answer

$\angle AOB + \angle COD = 180^\circ$