d) In the adjoining figure, $$ A P $$ and $$ B P $$ are the angular bisectors of $$ \angle B A D $$ and $$ \angle A B C $$ respectively. Show that $$ \angle A P B=1 $$ right angle. e) In the given figure, $$ A B C $$ is an equilateral triangle. If $$ A B \angle C E $$, prove that $$ \angle A C E=\angle E C D $$.

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To address your questions, let's break them down step by step.

### Part d)

We need to show that $$ \angle APB = 1 $$ right angle, given that $$ AP $$ and $$ BP $$ are the angular bisectors of $$ \angle BAD $$ and $$ \angle ABC $$ respectively.

1. **Understanding the Angles**:
   - Let $$ \angle BAD = x $$ and $$ \angle ABC = y $$.
   - Since $$ AP $$ is the angle bisector of $$ \angle BAD $$, we have:
     $$
     \angle BAP = \frac{x}{2}
     $$
     and
     $$
     \angle DAP = \frac{x}{2}.
     $$
   - Similarly, since $$ BP $$ is the angle bisector of $$ \angle ABC $$, we have:
     $$
     \angle ABP = \frac{y}{2}
     $$
     and
     $$
     \angle CBP = \frac{y}{2}.
     $$

2. **Finding $$ \angle APB $$**:
   - Now, we can find $$ \angle APB $$:
     $$
     \angle APB = \angle BAP + \angle ABP = \frac{x}{2} + \frac{y}{2} = \frac{x + y}{2}.
     $$

3. **Using the Angle Sum Property**:
   - Since $$ A, B, C, D $$ are points in the plane, we can use the fact that the sum of angles around point $$ P $$ must equal $$ 360^\circ $$. 
   - If we consider the angles around point $$ P $$, we have:
     $$
     \angle BAP + \angle ABP + \angle APD + \angle DPC = 360^\circ.
     $$
   - However, we are interested in $$ \angle APB $$ specifically.

4. **Conclusion**:
   - If $$ x + y = 180^\circ $$ (which is true if $$ A, B, C, D $$ are points on a straight line), then:
     $$
     \angle APB = \frac{x + y}{2} = \frac{180^\circ}{2} = 90^\circ.
     $$
   - Thus, $$ \angle APB = 1 $$ right angle.

### Part e)

We need to prove that $$ \angle ACE = \angle ECD $$ given that $$ ABC $$ is an equilateral triangle and $$ AB \parallel CE $$.

1. **Understanding the Configuration**:
   - Since $$ ABC $$ is an equilateral triangle, we have:
     $$
     \angle ABC = \angle ACB = \angle CAB = 60^\circ.
     $$
   - Given that $$ AB \parallel CE $$, we can use the properties of parallel lines and transversals.

2. **Using Alternate Interior Angles**:
   - Since $$ AB \parallel CE $$ and $$ AC $$ acts as a transversal, we have:
     $$
     \angle ACB = \angle ACE.
     $$
   - Therefore, we can write:
     $$
     \angle ACE = 60^\circ.
     $$

3. **Using Corresponding Angles**:
   - Similarly, since $$ AB \parallel CE $$ and $$ CD $$ acts as a transversal, we have:
     $$
     \angle ABC = \angle ECD.
     $$
   - Thus, we can write:
     $$
     \angle ECD = 60^\circ.
     $$

4. **Conclusion**:
   - Since both $$ \angle ACE $$ and $$ \angle ECD $$ equal $$ 60^\circ $$, we conclude that:
     $$
     \angle ACE = \angle ECD.
     $$

This completes the proofs for both parts d) and e).
For part d), if $$ \angle BAD $$ and $$ \angle ABC $$ are on a straight line, then $$ \angle APB $$ is 90 degrees, which is one right angle. For part e), in an equilateral triangle with a parallel line, the alternate interior angles are equal, so $$ \angle ACE = \angle ECD $$.

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