Yea I wanna say you have the right answer
Sorry just commenting for something
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).
we are supposed to find
Which of these properties is enough to prove that a given parallelogram is also a Rectangle?
As we know from the theorem, if the diagonals of a parallelogram are congruent then the parallelogram is a rectangle.
The other options The diagonals bisect each other is not sufficient because in parallelogram diagonals always gets bisected , parallelogram becomes rectangles only if both the diagonals are of same length.
In a parallelogram The opposite angles and opposite sides are always equal.
Hence the correct option is
The diagonals are congruent.