Given: ABCD is a trapezoid, m∠1=m∠2, m∠3=m∠4, AB = CD, AC ∩ BD = O Prove: m∠AOD = m∠BCD

Mathematics

1 answer:

LiRa [457]3 years ago

6 0

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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$\frac{4y}{9} - \frac{2y}{3} = 10 \\ = > \frac{4y - 6y}{9} = 10 \\ = > \frac{ - 2y}{9} = 10 \\ = > y = 10 \times \frac{9}{ - 2} \\ = > y = - 5 \times 9 \\ = > y = - 45$