Geometry

Here is the answer to your query.

Consider two ∆ABC and ∆PQR. In these two triangles ∠B = ∠Q = 90�, AB = PQ and AC = PR.

We can prove the R.H.S congruence rule i.e. to prove ∆ABC ≅ ∆PQR

We need the help of SSS congruence rule.

We have AB = PQ, and AC = PR

So, to prove ∆ABC ≅ ∆PQR in SSS congruence rule we just need to show BC = QR

Now, using Pythagoras theorems in ∆ABC and ∆PQR

Now, in ∆ABC and ∆PQR

AB = PQ, BC = QR, AC = PR

∴ ∆ABC ≅ ∆PQR [Using SSS congruence rule]

So, we have AB = PQ, AC = PR, ∠B = ∠Q = 90� and we have proved ∆ABC ≅ ∆PQR. This is proof of R.H.S. congruence rule.

Hope! This will help you.

Cheers!!!

It's actually angle B angle Y

pfft no, it's AC = DF

AC is congruent to DF.

B=Y

or

C=Z

Line segment BC is congruent to Line Segment YZ

Angle "A" is congruent to Angle "D"

AAS is equal to angle-angle-side, and is descriptive of a triangle. JKL and MNO would be the sides and angles of a triangle. The two sides must be congruent to the opposite angle.

True.

Only if the given angle is between the two sides will the two triangles guarantee to be congruent (SAS), unless the given angle is a right angle (90°) in which case you now have RHS (Right-angle, Hypotenuse, Side) which does guarantee congruence.

False

Bc= qr

false

k,l n;lm;m

The triangles must be congruent.

angle B angle Y

(Tested, correct)

Nicki is not the answer, just ignore that.

Nope Congruent - SSS Apex. You're welcome.

__ - __

AC = XZ

• = is the similar sign

True