Question

Which statements are correct? Select three options. â–łBCD is similar to â–łBSR. 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

Only a, b and, d are true.

Given to us,

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 DC and RS are parallel. width= Which statements are correct? Select three options.

a.)   △BCD is similar to △BSR.

b.)   BR/RD = BS/SC

c.)   If the ratio of BR to BD is Two-thirds, then it is possible that

      BS = 6 and BC  = 3.

d.)   (BR)(SC) = (RD)(BS)

e.)   BR/ RS = BS/SC

Angles on the same side of the transversal

1.   We know that the angles on the same side of the transversal are equal,  also, RS is parallel to DC. Thus,

∠BDC = ∠BRS,  

∠BCD = ∠BSR,

And we can write that ∠CBD = ∠CBD because they are the common angle of the two triangles.

Therefore, ΔBCD ~ ΔBSR, the using Angle, Angle Angle (AAA) property for similar triangles.

Angle, Angle, Angle (AAA) property for a similar triangle

2.   Now, As already discussed  ΔBCD ~ ΔBSR, using the Angle, Angle Angle (AAA) property for the similar triangle.

$\dfrac{BC}{BS} = \dfrac{BD}{BR}\\\\\\\dfrac{(BS + SC)}{BS} = \dfrac{(BR + RD)}{BR}\\\\\\1 + \dfrac{SC}{BS} = 1+\dfrac{RD}{BR}\\\\\\\dfrac{SC}{BS} = \dfrac{BR}{RD}$

Inverting both sides,

$\dfrac{BR}{RD} = \dfrac{BS}{SC}$

Therefore, $\dfrac{BR}{RD} = \dfrac{BS}{SC}$ is true.

3.   From the above observation, , we can write

$\dfrac{BR}{RD} = \dfrac{BS}{SC}\\\\BR\times SC = RD\times BR$

Therefore, the (BR)(SC) = (RD)(BR) is true.

Hence, only a, b and, d are true.

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