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Anton
11 days 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:
Zina [3.8K]11 days ago
7 0

Answer:

The correct statements are;

1) ΔBCD is similar to ΔBSR

2) BR/RD = BS/SC

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

Step-by-step explanation:

1) Since RS is parallel to DC, we conclude that;

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

Furthermore;

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

∠CBD = ∠CBD (Reflexive property)

Thus;

ΔBCD ~ ΔBSR by the Angle-Angle-Angle (AAA) similarity criterion.

2) Given that  ΔBCD ~ ΔBSR, we obtain;

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

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

SC/BS = RD/BR

By inverting both sides we find;

BR/RD = BS/SC

3) From BR/RD = BS/SC, we apply cross multiplication;

BR/RD = BS/SC leads to;

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

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