Hello, abrogard!

I'm not laughing . . . This was tricky.

There's a quadrilateral ABCD: $\displaystyle \angle B = 50^o,\;\;\angle D = 120^o$

Angles $\displaystyle BAD$ and $\displaystyle BCD$ are bisected. .The bisectors intersect at $\displaystyle T.$

Find $\displaystyle \angle ATC.$ Code:

A o * * * * * * o D
* * 120°
* * *
* *
* T o *
* *
* * *
* *
* * *
* *
* * *
* 50° *
B o * * * * * * * o C

The angles of a quadrilateral total 360°.

. . Hence: .$\displaystyle A + B + C + D \:=\:360^o$

Since $\displaystyle A = 50^o,\;D = 120^o$ we have: .$\displaystyle A + 50^o + C + 120^o \:=\:360^o \quad\Rightarrow\quad A + C \:=\:190^o$

. . Hence: .$\displaystyle \frac{1}{2}(A + C) \:=\:95^o$ .[1]

In quadrilateral $\displaystyle ADCT\!:\;\;\angle ATC + \angle TAD + \angle D + \angle DCT \:=\:360^o$

Since $\displaystyle \angle TAD = \frac{1}{2}A,\;\;\angle D = 120^o,\;\;\angle DCT = \frac{1}{2}C$

. . we have: .$\displaystyle \angle ATC + \frac{1}{2}A + 120^o + \frac{1}{2}C \:=\:360^o $

. . or: .$\displaystyle \angle ATC + \frac{1}{2}(A + C) \:=\:240^o $

Substitute [1]: .$\displaystyle \angle ATC + 95^o \:=\:240^o \quad\Rightarrow\quad\boxed{ \angle ATC \:=\:145^o }$