Solve the simultaneous equation, will give out brainliest answer a=bcd
a+b=cd
a+b+c=d
a+b+c+d=1​

a = 1/42, b = 1/7, c = 1/3 and d = 1/2

Step-by-step explanation:

a = bcd (1)

a + b = cd (2)

a + b + c = d (3)

a + b + c + d = 1  (4)​

Substituting equation (3) into equation (4), we have

a + b + c = d (3)

a + b + c + d = 1  (4)​

(a + b + c) + d = 1  (4)​

d + d = 1  (5)

2d = 1

dividing through by 2, we have

d = 1/2

Taking equation (2) and equation (3), we have

a + b = cd (2)

a + b + c = d (3)

substituting equation (2) into equation (3), we have

a + b = cd (2)

(a + b) + c = d (3)

cd + c = d

Factorizing, we have

c(d + 1) = d

dividing through by d + 1, we have

c = d/(d + 1)

substituting d = 1/2 into the equation, we have

c = d/(d + 1)

c = 1/2 ÷ (1/2 + 1)

c = 1/2 ÷ 3/2

c = 1/2 × 2/3

c = 1/3

Taking equations (1) and (2), we have

a = bcd (1)

a + b = cd (2)

Substituting equation (1) into (2), we have

a = bcd (1)

a + b = cd (2)

bcd + b = cd (2)

Factorizing, we have

b(cd + 1) = cd

dividing through by (cd + 1), we have

b = cd/(cd + 1)

substituting the values of c and d into the equation,, we have

b = cd/(cd + 1)

b = 1/3 × 1/2/(1/3 × 1/2 + 1)

b = 1/6/(1/6 + 1)

b = 1/6 ÷ 7/6

b = 1/6 × 6/7

b = 1/7

Since a = bcd, substituting the values of b,c and d into the equation, we have

a = bcd

a = 1/7 × 1/3 × 1/2

a = 1/42

So, a = 1/42, b = 1/7, c = 1/3 and d = 1/2

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

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