Given: δabc is isosceles; ab ≅ ac prove: ∠b ≅ ∠c we are given that δabc is isosceles with ab ≅ ac. using the definition of congruent line segments, we know that . let’s assume that angles b and c are not congruent. then one angle measure must be greater than the other. if m∠b is greater than m∠c, then ac is greater than ab by the . however, this contradicts the given information that . therefore, , which is what we wished to prove. similarly, if m∠b is less than m∠c, we would reach the contradiction that ab > ac. therefore, the angles must be congruent.

AB = AC

Triangle parts relationship theorem

side AB is congruent to side AC

angle B is congruent to angle c

Step-by-step explanation:

Step-by-step explanation:

Given: ΔABC is isosceles; AB ≅ AC

Prove: ∠B ≅ ∠C

Proof:

It is given that ΔABC is isosceles with AB ≅ AC. Using the definition of congruent line segments, we know that AB=AC.

Let us assume that angles B and C are not congruent. Then one angle measure must be greater than the other. If m∠B is greater than m∠C, then AC is greater than AB by the Triangle parts relationship theorem.

Now, this contradicts the given information that the side AB is congruent to the side AC that is AB ≅ AC.

Therefore, AB ≅ AC that we wished to prove.

Similarly, if m∠B is less than m∠C, we would reach the contradiction that AB > AC. Therefore, the angles must be congruent that is angle B is congruent to angle C.

Hence proved.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

w'x is 0.

hope it