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ABCD is a parallelogram. P is the midpoint ofside CD, seg BP meets diagonal AC at X. Prove that3AX = 2AC. |
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Answer» 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.. hope it heps u my friend!! |
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