Question

answer the following questions

Answer 1

hey here here is your answer in the pic please mark it as is brainliest hope it helps

Answer 2

Answer for Part 1

Given,

AB = CD

AB ║CD

To Prove,

ΔAOB ≅ ΔDOC

Proof

i)  AB = CD [Given]

  AB║CD and AD is the transversal.

∴ ∠BAO = ∠CDO [Alternate Interior Angles]

  AB║CD and BC is the transversal.

∴ ∠ABO = ∠DCO [Alternate Interior Angles]

∴ ΔAOB  ≅ ΔDOC by ASA Congruency

ii) AO = DO (CPCT)

iii) BO = CO (CPCT)

Hence Proved!

Answer for Part 2

In figure,

∠ECD = ∠BCA = 53° [V.O.A]

AB ║ DE and BD is the transversal.

→ ∠x = ∠D [Alternate interior angles]

→ ∠x = 45°

In ΔABC

→ ∠A  + ∠B + ∠C = 180° [A.S.P of a triangle]

→ ∠A + 45° + 53° = 180°

→ ∠A + 98° = 180°

→ ∠A = 180° - 98°

→ ∠A = 82°

AB║CD and AC is the transversal

[Alternate interior angles]

In Δ CDE

∠ECD + ∠CDE + ∠CED = 180° [A.S.P of a triangle]

53° + 45° + 3y + 1 = 180°

99° + 3y = 180°

3y = 180° - 99°

3y = 81°

y = $\frac{81}{3}$

y = 27°

[If you want to find the value of the 'angle' CED, you need to substitute the value of 'y' in 3y + 1]

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Therefore the values are

x = 45°

y = 27°

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Thank You!