In a `Delta ABC, AB=AC, D` is a point BC such that `angle BAD=50^(@)` and a point E on side AC such that AE =AD find `angle CDE`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have triangle ABC where AB = AC, which means triangle ABC is isosceles. We also know that angle BAD = 50° and that AE = AD. We are required to find angle CDE.

### Step 2: Draw the Triangle
Draw triangle ABC with points A, B, and C. Mark point D on line BC such that angle BAD = 50°. Then, mark point E on line AC such that AE = AD.

### Step 3: Identify Angles
Since AB = AC, angle ABC = angle ACB. Let's denote:
- angle BAD = 50° (given)
- angle ADB = angle AEB (since AE = AD)

Let:
- angle BAD = angle 1
- angle ADB = angle 2
- angle ACB = angle 3

### Step 4: Apply Angle Sum Property in Triangle ADE
In triangle ADE, the sum of angles is 180°:
\[ \text{angle ADE} + \text{angle ADB} + \text{angle DAE} = 180° \]
This can be expressed as:
\[ 2 \cdot \text{angle 1} + \text{angle 3} = 180° \]
(Here, angle DAE = angle ADB = angle 1)

### Step 5: Apply Angle Sum Property in Triangle ABC
In triangle ABC, the sum of angles is also 180°:
\[ \text{angle ABC} + \text{angle ACB} + \text{angle BAC} = 180° \]
This can be expressed as:
\[ 2 \cdot \text{angle 2} + \text{angle 3} + 50° = 180° \]

### Step 6: Set Up the Equations
From triangle ADE:
\[ 2 \cdot \text{angle 1} + \text{angle 3} = 180° \quad \text{(Equation 1)} \]

From triangle ABC:
\[ 2 \cdot \text{angle 2} + \text{angle 3} + 50° = 180° \quad \text{(Equation 2)} \]

### Step 7: Solve the Equations
From Equation 2:
\[ 2 \cdot \text{angle 2} + \text{angle 3} = 130° \quad \text{(Rearranging)} \]

Now, we have:
1. \( 2 \cdot \text{angle 1} + \text{angle 3} = 180° \)
2. \( 2 \cdot \text{angle 2} + \text{angle 3} = 130° \)

Subtract Equation 2 from Equation 1:
\[ (2 \cdot \text{angle 1} - 2 \cdot \text{angle 2}) = 50° \]
Thus:
\[ \text{angle 1} - \text{angle 2} = 25° \]

### Step 8: Find angle CDE
Now, we need to find angle CDE. By the exterior angle theorem in triangle CDE:
\[ \text{angle CDE} + \text{angle 2} = \text{angle 1} \]
Therefore:
\[ \text{angle CDE} = \text{angle 1} - \text{angle 2} = 25° \]

### Final Answer
Thus, angle CDE = 25°.

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