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aalyn [17]
2 years ago
13

Line RS intersects triangle BCD at two points and is parallel to segment DC. Triangle B C D is cut by line R S. Line R S goes th

rough sides D B and C B. Lines D C and R S are parallel. width= Which statements are correct? Select three options. △BCD is similar to △BSR. StartFraction B R Over R D EndFraction = StartFraction B S Over S C EndFraction If the ratio of BR to BD is Two-thirds, then it is possible that BS = 6 and BC = 3. (BR)(SC) = (RD)(BS) StartFraction B R Over R S EndFraction = StartFraction B S Over S C EndFraction

Mathematics
1 answer:
Svetlanka [38]2 years ago
7 0

Answer:

The correct options are;

1) ΔBCD is similar to ΔBSR

2) BR/RD = BS/SC

3) (BR)(SC) = (RD)(BS)

Step-by-step explanation:

1) Given that RS is parallel to DC, we have;

∠BDC = ∠BRS (Angles on the same side of transversal)

Similarly;

∠BCD = ∠BSR (Angles on the same side of transversal)

∠CBD = ∠CBD = (Reflexive property)

Therefore;

ΔBCD ~ ΔBSR Angle, Angle Angle (AAA) rule of congruency

2) Whereby  ΔBCD ~ ΔBSR, we therefore have;

BC/BS = BD/BR → (BS + SC)/BS = (BR + RD)/BR = 1 + SC/BS = RD/BR + 1

1 + SC/BS = 1 + RD/BR = SC/BS = 1 + BR/RD - 1

SC/BS = RD/BR

Inverting both sides

BR/RD = BS/SC

3) From BR/RD = BS/SC the above we have by cross multiplication;

BR/RD = BS/SC gives;

BR × SC = RD × BR → (BR)(SC) = (RD)(BR).

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1. Solve for w in the equation W/5 = 25<br> A.W = 5<br> B.W = 125<br> C. W = 30<br> D.W = 20
vekshin1

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