Find the side of a convex four sided polygon that extends to a triangle

Question

Convex polygon $ABCD$ of four sides has angles $\alpha+\gamma=\beta+\delta=180°$.

Extend $CD$ and $BA$ to intersect at $T$. Then, $TD$ is $6$ cm, $TC$ is $10$ cm, and $TA$ is $3$ cm.

Find $AB$.

Image of the problem.

---

The solution is $17$ cm, but I do not know how to arrive at that conclusion.

I trivially know that angle $ADT$ equals $\beta$ and that $TAD$ equals $\gamma$.

Answer 1

Let $AB=x$.

Since $ABCD$ is cyclic, we obtain: $3(3+x)=6\cdot10$, which gives $x=17$.

Answer 2

Triangles $TAD$ and $TCB$ are similar. So $$ AB=TB-TA=TD\cdot\frac{TC}{TA}-TA$$

Answer 3

Considering triangle TDA: $$\frac{TD}{\sin(180-\alpha)} = \frac{TC}{\sin(\beta)}$$ $$\Longleftrightarrow \frac{\sin(\alpha)}{\sin(\beta)} =2 $$

Considering triangle TCB: $$ \frac{TD}{\sin(180-\alpha)} = \frac{TA}{\sin(\beta)}$$ $$\Longleftrightarrow TB = 10\frac{\sin(\alpha)}{\sin(\beta)} = 20 $$ $$\Longrightarrow AB=17$$

Answer 4

The opposite angles are supplementary. This is a characteristic of cyclic quadrilaterals or quadrilaterals which are able to be circumscribed. Because of this we can obtain that lines TB and TC are secants to this circle. A common theorem relating two secants in a circle is, in this specific scenario,

TA(TB)=TD(TC)

Give this formula a try and you'll get the answer lickity split.