ABCD is a parallelogram P is a mid-point of the side CD. BP meets the diagonal AC at X. Prove that AX=2 CX
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Answer:
2AC=3AX
Step-by-step explanation:
in triangle ABX and triangle CPX angle ABX=angle CPX.......(alternate angles)
angle BAX=angle PCX.....(alternate angles)
Therefore triangle ABX is similar to triangle CPX by AA test of similarity
AB/CP=AX/CX.......corresponding sides of similar triangle
AB=2PC.....(AB=CD...opposite side if parallelogram and P is the midpoint of side CD)
AX+XC=AC XC=AC-AX
2PC/PC=AX/AC-AX.....(substitute the values)
2AC-2AX=AX
2AC=2AX+AX
2AC=3AX
Hence proved
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