Complete the paragraph proof. given: ∠abr and ∠acr are right angles ab ≅ bc bc ≅ ac prove: bisects ∠bac it is given that ∠abr and ∠acr are right angles, ab ≅ bc and bc ≅ ac since they contain right angles, △abr and △acr are right triangles. the right triangles share hypotenuse ar, and reflexive property justifies that ar ≅ ar. since ab ≅ bc and bc ≅ ac, the transitive property justifies ab ≅ ac. now, the hypotenuse and leg of right △abr is congruent to the hypotenuse and the leg of right △acr, so △abr ≅ △acr by the hl congruence postulate. therefore, by cpctc, and bisects ∠bac by the definition of bisector.

The answer is <BAR ≅<CAR :)

∠BAR=∠CAR [∴ΔABR≅ΔACR]

Step-by-step explanation:

In ΔABR and ΔACR

AB=AC=X [ ∴AB≅BC, and BC≅AC, So AB≅AC]

∠ABR = ∠ACR [ each being 90°]

AR is common.

ΔABR ≅ ΔACR [ RHS]

RHS means if in two right triangles hypotenuse and one side of a triangle is equal to hypotenuse and other side then the two triangles are congruent.

So ,∠BAC is bisected.

i.e, ∠BAR=∠CAR [ CPCT ]

Hence proved.

step-by-step explanation:

the polynomial is a perfect square trinomial of the form a^2-2ab+b^2:

option a. true

step-by-step explanation:

9x^2-12x+4

comparing with the form a^2-2ab+b^2

1.) a^2=9x^2

solving for a: square root both sides of the equation:

sqrt(a^2)=sqrt(9x^2)

a=sqrt(9) sqrt(x^2)

a=3x

2.) b^2=4

solving for b: square root both sides of the equation:

sqrt(b^2)=sqrt(4)

b=2

to be a perfet square trinomial 2ab must be equal to 12x. let's check:

2ab = 2 (3x) (2)

2ab = 12x

correct