Question

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

Answer 1

3AX = 2AC

Step-by-step explanation:

Quadrilateral ABCD is a parallelogram. P is the midpoint of side CD

Lets draw XY ║AB ║ CD

=> XY ║ CP

=> ΔBXY ≈ ΔBPC

=> BX/BP = BY/BC = XY/PC

XY ║ AB

=> ΔCXY ≈ ΔCAB

=> CX/CA  = CY/CB = XY/AB

 (BY/BC)/(CY/CB)  = (XY/PC)/(XY/AB)

=> BY/CY  = AB/PC

PC = CD/2 = AB/2

=> BY/CY = 2

=> BY = 2CY

BC = BY + CY  = 2CY + CY = 3CY

BY = 2CY

CX/CA  = CY/CB

=> CX/CA  = CY/3CY

=> CX/CA = 1/3

=> CA = 3CX

AX = CA - (CA/3)

=> 3AX = 3CA - CA

=> 3AX = 2CA

=> 3AX = 2AC

QED

Proved

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