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Shalnov [3]
3 years ago
14

HELP!

Mathematics
2 answers:
astraxan [27]3 years ago
8 0

Answer:


Step-by-step explanation:

The two lines are parallel in the triangle if they satisfy the basic proportionality theorem that  is:

\frac{MX}{XL}=\frac{MY}{YN}

Now, different segments lengths are given as:

(A) LX=2, XM=7, NY=4 and YM=14

Applying the basic proportionality theorem,

\frac{7}{2}=\frac{14}{4}

\frac{7}{2}=\frac{7}{2}

Hence, The line segment LN is parallel to XY.

(B) LX=2, XM=6, NY=3 and YM=9

Applying the basic proportionality theorem,

\frac{6}{2}=\frac{9}{3}

\frac{3}{1}=\frac{3}{1}

Hence, The line segment LN is parallel to XY.

(C) LX=2, XM=3, NY=4 and YM=7

Applying the basic proportionality theorem,

\frac{3}{2}{\neq}\frac{7}{4}

Hence, The line segment LN is not parallel to XY.

Korvikt [17]3 years ago
8 0

Just took the test and got the the correct answer *\(♡°▽°♡)/* Look at the image down below!!

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Δ QRS ≈ Δ QST ≈ Δ SRT ⇒ 3rd answer

Step-by-step explanation:

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- From (1) and (3)

∴ m∠QST + m∠RST = m∠Q + m∠QST

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In Δ SRT

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- By using the fact above

∴ m∠R + m∠RST = 90 ⇒ (4)

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In Δs QRS and QST

∵ m∠S = m∠QTS ⇒ right angles

∵ m∠R = m∠QST ⇒ proved

∵ ∠Q is a common angle in the two Δs

∴ Δ QRS ≈ Δ QST ⇒ AAA postulate of similarity

In Δs QRS and SRT

∵ m∠S = m∠STR ⇒ right angles

∵ m∠Q = m∠RST ⇒ proved

∵ ∠R is a common angle in the two Δs

∴ Δ QRS ≈ Δ SRT ⇒ AAA postulate of similarity

If two triangles are similar to one triangle, then the 3 triangles are similar

∵ Δ QRS ≈ Δ QST

∵ Δ QRS ≈ Δ SRT

∴ Δ QRS ≈ Δ QST ≈ Δ SRT

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