given: (overline{ba}perpoverline{pa}); (overline{cd}perpoverline{pd}); point (p) is the midpoint of (overline{bc}). prove: (overline{bc}) bisects (overline{ad}). 4. (angle apbcongangle dpc) 5. point (p) is the midpoint of (overline{bc}). 6. (overline{ap}congoverline{pd})

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Accepted Answer

# Explanation:
## Step1: Prove $\triangle BAP\cong\triangle CDP$
- Since $BA\perp PA$ and $CD\perp PD$, $\angle BAP = \angle CDP=90^{\circ}$
- Point $P$ is the mid - point of $BC$, so $BP = CP$
- $\angle APB=\angle DPC$ (vertically opposite angles)
- By $AAS$ (Angle - Angle - Side) congruence criterion, $\triangle BAP\cong\triangle CDP$

## Step2: Use CPCTC
- If $\triangle BAP\cong\triangle CDP$, then by Corresponding Parts of Congruent Triangles are Congruent (CPCTC), $AP = PD$

## Step3: Use the definition of a bisector
- If $AP = PD$, then $P$ is the mid - point of $AD$
- Since $P$ lies on $BC$, $BC$ bisects $AD$

# Answer:
To prove $BC$ bisects $AD$, we first show $\triangle BAP\cong\triangle CDP$ using $AAS$ ( $\angle BAP=\angle CDP = 90^{\circ}$, $BP = CP$, $\angle APB=\angle DPC$). Then by CPCTC, $AP = PD$. Since $P$ is on $BC$ and $AP = PD$, $BC$ bisects $AD$

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Explanation:

Step1: Prove \(\triangle BAP\cong\triangle CDP\)

Step2: Use CPCTC

Step3: Use the definition of a bisector

Answer: