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abd=bad=70 adb=40 bca=20 and bdc=90 we need to find acd=x from this picture of gemoteric


From the image and the information provided, we are dealing with a quadrilateral \(ABCD\) that seems to have some diagonals intersecting at a certain point, creating various triangles within the quadrilateral. The known angles are:

- \(\angle ABD = \angle BAD = 70^\circ\)
- \(\angle ADB = 40^\circ\)
- \(\angle BCA = 20^\circ\)
- \(\angle BDC = 90^\circ\)
- We need to find \(\angle ACD = x^\circ\)

### Step-by-step approach:
1. **Triangle \(ABD\)**: 
   - Since \(\angle ABD = \angle BAD = 70^\circ\) and \(\angle ADB = 40^\circ\), we can conclude that triangle \(ABD\) is isosceles with \(\angle ADB = 40^\circ\).
   - The sum of angles in triangle \(ABD\) is \(180^\circ\), so we can find \(\angle BDA\) as:
     \[
     \angle BDA = 180^\circ - 70^\circ - 70^\circ = 40^\circ
     \]

2. **Triangle \(BDC\)**:
   - We are given that \(\angle BDC = 90^\circ\), making it a right triangle.
   - \(\angle BCA = 20^\circ\) is also given.

3. **Finding \(\angle ACD = x\)**:
   - Since quadrilateral \(ABCD\) has the diagonal \(AC\) and is not cyclic (based on the angle values), the sum of the angles in any triangle within the quadrilateral should be \(180^\circ\).
   - Using the known angles in triangle \(ACD\), we have:
     \[
     \angle ACD + \angle BDC + \angle BCA = 180^\circ
     \]
     Substituting the known values:
     \[
     x + 90^\circ + 20^\circ = 180^\circ
     \]
     \[
     x + 110^\circ = 180^\circ
     \]
     \[
     x = 180^\circ - 110^\circ = 70^\circ
     \]

Thus, \(\angle ACD = x = 70^\circ\).

Let me know if you need further clarifications or more details!

### Follow-up Questions:
1. How does the property of isosceles triangles help in solving this problem?
2. What are the conditions for a quadrilateral to be cyclic, and how does it affect angle relations?
3. Can you explain why \(\triangle BDC\) being a right triangle is significant in this case?
4. How can we use the exterior angle theorem in geometry to solve similar problems?
5. What are other ways of finding unknown angles in quadrilaterals based on diagonal intersections?

### Tip:
Always check for symmetry or special triangles (like isosceles or right triangles) in geometry problems, as they often simplify the process of finding unknown angles.

abd=bad=70 adb=40 bca=20 and bdc=90 we need to find acd=x from this picture of gemoteric

Learn how to solve for the unknown angle ACD in quadrilateral ABCD using properties of isosceles triangles and the angle sum theorem. With given angles of ABD = BAD = 70°, ADB = 40°, BCA = 20°, and BDC = 90°, find ACD step-by-step. Suitable for students in Grades 9-10.

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