Question

ABCD is a parallelogram. P is the midpoint of side CD.
Seg BP meets diagonal AC at X. Prove that 3AX = 2AC.​

Answer 1

Step-by-step explanation:

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Answer 2

Given:

• ABCD is a parallelogram
• BP meets the diagonal AC at X.
• P is the midpoint of the side CD.

To Find:

• To Prove: 3AX = 2AC

Solution:

• First, let us construct a diagram using the given data.
• From the diagram, we can come to few conclusions like:
• Consider ΔABC and ΔCPX, AB is paralell to CD
• ∠BAX = ∠CPX (∵ Alternate angles)
• ∠BXA = ∠PXC (∵ Vertically opposite angles)
• From the above conclusions, we get to know that  ΔABx ≅ ΔCPX
• We know that, when two triangles are similar, their corresponding sides are equal.
• $\frac{AX}{CX} = \frac{AB}{CP}$ ⇒ $\frac{AX}{AC-AX} = \frac{2CP}{CP}$ (∵ since p is the midpoint of CD)
• $\frac{AX}{AC-AX} = 2$ ⇒ AX = 2AC-2AX
• 3AX = 2AC

Hence proved.

3AX=2AC