Question

ABCD is a rectangle. M is the mid-point of AC and CPMN is a rectangle. Prove that:

1. P is the mid-point of CD
2. PN = `1/2` AC

Answer 1

Given:

• ABCD is a rectangle.
• M is the mid-point of AC.
• CPMN is a rectangle.

To Prove:

1. P is the mid-point of CD. 
2. PN = `1/2` AC.

Proof:

Step 1: Since ABCD is a rectangle

• AB = CD and AD = BC,
• All angles are 90°.

Step 2: M is the mid-point of diagonal AC, So, AM = MC = `1/2` AC.

Step 3: CPMN is a rectangle

• Opposite sides are equal and parallel,
• All angles are 90°,
• Therefore, CP = MN and CM = PN,
• CP || MN and CM || PN.

Step 4: Since CPMN is a rectangle and shares side CM with ABCD and M is midpoint of AC, triangle AMC is formed.

Step 5: Since CPMN is a rectangle and P, C, M, N are its vertices,

• P lies on CD,
• N lies on AB,
• Because CPMN is rectangle, sides CP and MN are parallel and sides CM and PN are parallel,
• Since M is midpoint of AC and C and P lie on CD,
• By midpoint theorem in triangle ADC, P is midpoint of CD.

Step 6: Since CPMN is rectangle

• PN = CM (opposite sides of rectangle)
• CM = MA (since M is midpoint of AC)
• PN = `1/2` AC