Question

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 through 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

Answer 1

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).

Answer 2

This question is related to properties of similar figures. Therefore, the three statements that are correct are:

• ΔBCD is similar to ΔBSR  
• $\frac{BR}{RD}$ = $\frac{BS}{SC}$  
• (BR)(SC) = (RD)(BS)

Two or more figures or shapes are said to be similar if they have all properties in common. This implies angles, orientation and parallel/ perpendicular sides etc.

Comparing the properties of the sketch for the question, the required statements that are correct are:

a. ΔBCD is similar to ΔBSR  

b. $\frac{BR}{RD}$ = $\frac{BS}{SC}$

c. (BR)(SC) = (RD)(BS)

The sketch for the question is attached for better comparison.

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