Prove that a line that divides two sides of a triangle proportionally is parallel to the third side. Be sure to create and name
the appropriate geometric figures.
1 answer:
Answer:
Given: A triangle ABC and a line DE parallel to BC.
To prove: A line parallel to one side of a triangle divides the other two sides proportionally.
Proof: Consider ΔABC and DE be the line parallel to Bc, then from ΔABC and ΔADE, we have
∠A=∠A (Common)
∠ADE=∠ABC (Corresponding angles)
Thus, by AA similarity, ΔABC is similar to ΔADE, therefore
AB/AD= AC/AE
⇒AD+DB/AD = AE+EC/AE
⇒1+DB/AD = 1+ EC/AE
⇒DB/AD = EC/AE
Therefore, a line parallel to one side of a triangle divides the other two sides proportionally.
⇒Therefore Proved
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Answer:
The values of x and y in the diagonals of the parallelogram are x=0 and y=5
Step-by-step explanation:
Given that ABCD is a parallelogram
And segment AC=4x+10
From the figure we have the diagonals AC=3x+y and BD=2x+y
By the property of parallelogram the diagonals are congruent
∴ we can equate the diagonals AC=BD
That is 3x+y=2x+y
3x+y-(2x+y)=2x+y-(2x+y)
3x+y-2x-y=2x+y-2x-y
x+0=0 ( by adding the like terms )
∴ x=0
Given that segment AC=4x+10
Substitute x=0 we have AC=4(0)+10
=0+10
=10
∴ AC=10
Now (3x+y)+(2x+y)=10
5x+2y=10
Substitute x=0, 5(0)+2y=10
2y=10
∴ y=5
∴ the values of x and y are x=0 and y=5