1. ## is this wrong?

A,E,D are collinear.
m(ACB)=50
m(ABC)=40
m(CBD)=70
m(BCD)=65
m(BDA)=?

2. ## Re: is this wrong?

I do not understand your notation. What is the $m$ function, how does it turn three letters into number? Is it the measure of the angle in degrees? Is it the area of the triangle formed by the three vertices? Need a little more here.

3. ## Re: is this wrong?

Angle ACB.It is not a function. They are angle degrees.

4. ## Re: is this wrong?

Do you know anything about $m(\angle AEB), m(\angle AEC), m(\angle BED),\text{ or } m(\angle CED)$? So far, I know that $m(\angle BAC) = 90^\circ$ and $m(\angle BDC) = 45^\circ$. There is not enough information to know $m(\angle BDA)$.

6. ## Re: is this wrong?

Originally Posted by kastamonu
Can you translate? is that a kite?

7. ## Re: is this wrong?

No it is not a kite.
A,E,D are collinear.
Angle(ACB)=50 degrees
Angle(ABC)=40 degrees
Angle(CBD)=70 degrees
Angle(BCD)=65 degrees
m(BDA)=?

8. ## Re: is this wrong?

According to the messages there is an angel bisector but I couldn't see it.The question is ok. I translated all.

9. ## Re: is this wrong?

Maybe we must use the incredible trio.There is a right angle there.

10. ## Re: is this wrong?

Originally Posted by kastamonu
Maybe we must use the incredible trio.There is a right angle there.
It is very clear that$\angle BAC$ is a right angle.
But I agree with Slip, there is not sufficient information to answer the question.
Is $\overline{AD}$ an angular bisector?

11. ## Re: is this wrong?

Did it this way:
Assign value 1 to AB.
Use Sine Law 3 times.
Use Cosine law once.

12. ## Re: is this wrong?

We can draw a median through A. A is a right angle. We will get isosceles triangles by this way. Answer is 30 degrees.

13. ## Re: is this wrong?

You're dreaming in Technicolor!
Stop making wild guesses and start drawing
your diagrams to scale: what you're showing is a mess!
Drawing a median from A makes no sense.