1. ## Find An Angle in A Quadrilateral.

I have this problem I can't figure out. Hopefully someone can help me.

There's a quadrilateral ABCD. The angle ABC is 50degrees. The angle ADC is 120 degrees.

The angle BAD is bisected and the angle BCD is bisected. These lines of bisection meet at a point T.

What is the angle ATC?

Please don't laugh at how dumb I am. I know.

regards,

ab

2. Hello, abrogard!

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

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

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

Find $\angle ATC.$
Code:
              A o * * * * * * o D
* *         120°
*   *            *
*     *
*     T o           *
*          *
*             *        *
*                *
*                   *     *
*                      *
*                         *  *
* 50°                        *
B o   *   *   *   *   *   *   *   o C
The angles of a quadrilateral total 360°.
. . Hence: . $A + B + C + D \:=\:360^o$

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

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

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

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

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

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

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

3. Thank you Soroban. It dawned on me in the early hours this morning. Making me wonder how I could be so dumb. It was my mind insisting it had to know the absolute value of the angles at A and C, bisected or not, rather than simply deal with the total they represent of the original quadrilateral and the newly formed one.

Preconceptions preventing me finding an answer. Stubbornly sticking to a fallacy.

It's amusing, eh?

thanks again.

regards,

ab