Given: ABCD is a trapezoid, m∠1=m∠2, m∠3=m∠4, AB = CD, AC ∩ BD = O
Prove: m∠AOD = m∠BCD
1 answer:
Since AB = CD the trapezoid is isosceles, which means that ∡A = ∡D Therefore also ∡2 = ∡3 (they are half of the congruent angles) For the properties of parallel lines (BD and AD) crossed by a transversal (BD) we have ∡3 = ∡CBD. Now consider triangles AOD and BCD: ∡OAD (2) = ∡ADO (3) = ∡CBD (3) = ∡CDB (4) T<span>he sum of the angles of a triangle must be 180°, t</span>herefore: ∡AOD = 180 - ∡2 - ∡3 ∡BCD = 180 - ∡3 - ∡4 ∡AOD = ∡BCD because their measure is the difference of congruent angles.
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Answer:
a = 5
Step-by-step explanation:
a +a+4 + 2a-3 = 21
4a +1 =21
4a = 20
a = 5
Let, the numbers = x,y
x+y= 24 1st eqn
x-y = 44 2nd eqn
24-y-y = 44
-2y = 44-24
y = 20/-2 = -10
substitute that in equation 1st x = 24+10 = 34
so, the numbers would be 34 & -10
C. 0.4 times 6 is 2.4. Hope that helps.
Annual = $68,000 Semiannual = $34,000 Monthly = $5,666.67 Semimonthly or bimonthly = $2,833.33 Weekly = $1,416.67
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