In this problem, we're going to explain why any trapezoid that can be inscribed in a circle must be an isosceles trapezoid. in this figure, let's assume without loss of generality that segments ab and dc are parallel. by the end of this problem, we want to show that ∠d ≅ ∠c.

1. explain why ∠a must be supplementary to ∠d.

2. explain why ∠a must also be supplementary to ∠c.

3. now we know that ∠a is supplementary to ∠d and is also supplementary to ∠c. explain why that proves that ∠d must be congruent to ∠c.

1. By co-interior angles

2.by property of cyclic trapezoid.

Step-by-step explanation:

Given a trapezoid that can be inscribed in a circle. we have to prove ∠D ≅ ∠C. by the end of problem.

As ABCD is a trapezoid.

⇒ AB||DC

Since, AB||DC ∴ ∠BAD and ∠CDA are the co-interior angles. As a result, the sum of above two is 180° i.e these two are supplementary.

Hence, ∠A must be supplementary to ∠D.

2. As the trapezoid is inscribed in a circle which means trapezoid is cyclic.

⇒ The sum of the opposite angles of a cyclic trapezoid is supplementary.

Hence, ∠A must also be supplementary to ∠C.

Now, ∠A is supplementary to ∠D and is also supplementary to ∠C.

i.e ∠A + ∠D = 180°  →   (1)

   ∠A + ∠C=180°    →   (2)

From (1) and (2), we get

∠A + ∠D = ∠A + ∠C

⇒ ∠D = ∠C

Hence, ∠D must be congruent to ∠C.

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

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step-by-step explanation: