BD is the disector of angle ABC. From a point P in BD, perpendiculars PE and PF are drawn to AB and BC respectively, prove that :
(i) Triangle BEP is conguent to triangle BFP (ii) PE=PF.

Answer

To solve the problem step by step, let's break it down into two parts as required by the question.

### Given:
- BD is the bisector of angle ABC.
- P is a point on BD.
- PE and PF are perpendiculars drawn from P to AB and BC respectively.

### To Prove:
1. Triangle BEP is congruent to triangle BFP.
2. PE = PF.

### Step-by-Step Solution:

#### Part (i): Proving Triangle BEP ≅ Triangle BFP

1. **Identify the Angles**:
- Since BD is the bisector of angle ABC, we have:
\[ \angle ABP = \angle CBP \quad \text{(let's call this angle 1 and angle 2)} \]

2. **Identify Right Angles**:
- Since PE is perpendicular to AB, we have:
\[ \angle PEB = 90^\circ \]
- Since PF is perpendicular to BC, we have:
\[ \angle PFB = 90^\circ \]

3. **Common Side**:
- The side BP is common to both triangles BEP and BFP:
\[ BP = BP \]

4. **Apply the Angle-Side-Angle (ASA) Congruence Criterion**:
- We have:
- \(\angle ABP = \angle CBP\) (Angle 1 = Angle 2)
- \(\angle PEB = \angle PFB = 90^\circ\)
- \(BP = BP\) (Common side)
- Therefore, by the ASA criterion:
\[ \triangle BEP \cong \triangle BFP \]

#### Part (ii): Proving PE = PF

5. **Using Corresponding Parts of Congruent Triangles**:
- Since we have established that \(\triangle BEP \cong \triangle BFP\), by the property of congruent triangles, we can say:
\[ PE = PF \quad \text{(CPCT)} \]

### Conclusion:
1. We have proved that triangle BEP is congruent to triangle BFP.
2. We have also proved that PE = PF.