It is given that and are right angles, and . Since they contain right angles, ΔABR and ΔACR are right triangles. The right triangles share hypotenuse , and reflexive property justifies that . Since and , the transitive property justifies . Now, the hypotenuse and leg of right ΔABR is congruent to the hypotenuse and the leg of right ΔACR, so by the HL congruence postulate. Therefore, by CPCTC, and bisects by the definition of bisector.

THIS IS THE COMPLETE QUESTION BELOW;

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

Step-by-step explanation:

From the question, In the ΔABR and ΔACR , AB=AC=X

CHECK THE ATTACHMENT FOR DETAILED STEP BY STEP EXPLANATION:

A

Step-by-step explanation:

I hope this helps.

757764

step-by-step explanation:

de77737dd

irrational, then, just means all the numbers that aren't rational. every integer is a rational number, since each integer n can be written in the form n/1. however, numbers like 1/2, 45454737/2424242, and -3/7 are also rational, since they are fractions whose numerator and denominator are integers.

step-by-step explanation: