Question

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

Answer 1

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Answer 2

Step-by-step explanation:

Given : ABCD is a parallelogram of which sides AB and CD are parallel.

            P is the midpoint of the side CD.

To prove : 3AX = 2AC

Solution : Since AB ║ CD and AC is a transverse then

               ∠PBA ≅ ∠BPC [ Alternate angles ]

Similarly AB ║ CD and BP is the transverse then

               ∠ACD ≅ ∠CAB [Alternate angles]

By AA theorem of similarity ΔAXB and ΔPXC will be similar.

Now we know in two similar triangles their corresponding sides will be in the same ratio.

Therefore, $\frac{AX}{CX}=\frac{AB}{PC}$

Since CX = AC - AX and side DP = side PC = $\frac{1}{2}AB$

Now the ratio of the corresponding sides will be,

$\frac{AX}{AC-AX}=\frac{AB}{\frac{1}{2}AB}$

$\frac{AX}{AC-AX}=2$

AX = 2(AC - AX)

AX + 2AX = 2AC

3AX = 2AC

Hence proved.

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