In `DeltaABC,angleB=90^(@) and AE and CF` are the angle bisector (line segments that divide an angle into two equal parts) of `angleA and angleC`, respectively. Find `angleADC`.

Answer

To find the angle \( \angle ADC \) in triangle \( ABC \) where \( \angle B = 90^\circ \) and \( AE \) and \( CF \) are the angle bisectors of \( \angle A \) and \( \angle C \) respectively, we can follow these steps:

### Step 1: Understand the Properties of Triangle ABC
In triangle \( ABC \):
- \( \angle B = 90^\circ \)
- The sum of angles in a triangle is \( 180^\circ \).

### Step 2: Use the Angle Sum Property
Using the angle sum property of triangle \( ABC \):
\[ \angle A + \angle B + \angle C = 180^\circ \]
Substituting \( \angle B = 90^\circ \):
\[ \angle A + 90^\circ + \angle C = 180^\circ \]
This simplifies to:
\[ \angle A + \angle C = 90^\circ \]

### Step 3: Define the Angles Created by the Angle Bisectors
Let:
- \( \angle A = 1 + 2 \) (where \( AE \) is the angle bisector, so \( 1 = 2 \))
- \( \angle C = 3 + 4 \) (where \( CF \) is the angle bisector, so \( 3 = 4 \))

From the angle bisector properties, we have:
\[ 1 = 2 \quad \text{and} \quad 3 = 4 \]

### Step 4: Substitute the Angle Bisector Values
Since \( \angle A + \angle C = 90^\circ \), we can write:
\[ (1 + 1) + (3 + 3) = 90^\circ \]
This simplifies to:
\[ 2 \cdot 1 + 2 \cdot 3 = 90^\circ \]
Thus:
\[ 2(1 + 3) = 90^\circ \]
Dividing by 2:
\[ 1 + 3 = 45^\circ \]
So, we conclude:
\[ \angle 1 + \angle 3 = 45^\circ \]

### Step 5: Apply the Angle Sum Property in Triangle ADC
In triangle \( ADC \):
\[ \angle 2 + \angle 4 + \angle ADC = 180^\circ \]
Substituting \( \angle 2 = 1 \) and \( \angle 4 = 3 \):
\[ 1 + 3 + \angle ADC = 180^\circ \]
From our previous result, we know \( 1 + 3 = 45^\circ \):
\[ 45^\circ + \angle ADC = 180^\circ \]

### Step 6: Solve for Angle ADC
Now, isolate \( \angle ADC \):
\[ \angle ADC = 180^\circ - 45^\circ = 135^\circ \]

### Final Answer
Thus, the measure of \( \angle ADC \) is:
\[ \angle ADC = 135^\circ \]

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