(a) ∠AEC = ∠CFD, by transitive property of equality
(b) ∠CDE = 108°, ∠DCB = 108°, ∠ABC = 108°,∠EAB = 144°, ∠AED = 72°
The reason the above values are correct is as follows:
(a) The given parameters are;
Figure ABCDE is a pentagon
The sides AE is parallel to side CD
Line EC is parallel to side BC
The point of intersection of the extension off side ED and BC = Point F
∠ABC = ∠CDE
Required:
To show that ∠AEC = ∠CFD
Method:
Draw the pentagon ABCDE and include the added construction
Analyze the drawing
Solution:
∠ECF and ∠ABC are corresponding angles between parallel lines EC ║BC
∴ ∠ECF ≅ ∠ABC by corresponding angles formed by parallel lines are congruent
∠ECF = ∠ABC by definition of congruency
∠ABC = ∠CDE = ∠ECF by transitive property
∠ECD ≅ ∠AEC by alternate angles formed between parallel lines having a common transversal
∠ECD = ∠AEC by definition of congruency
∠ECF = ∠FCD + ∠ECD by angle addition postulate
∠CDE = ∠FCD + ∠CFD by exterior angle theorem
From ∠CDE = ∠ECF above, we have;
∠ECF = ∠FCD + ∠ECD = ∠FCD + ∠CFD
∴ ∠ECD = ∠CFD by addition property
∠ECD = ∠AEC, therefore;
∠AEC = ∠CFD, by transitive property
(b) Given that ΔEDC and ΔDFC are both isosceles triangles, with sides;
ED = DC, and DF = FC;
Let r represent ∠CFD, we have;
∠FCD = ∠CDF by base angles of isosceles triangle ΔDFC
∠ECD = ∠CED by base angles of isosceles triangle ΔEDC
∠AEG = ∠CDE by corresponding angles formed by parallel lines having a common transversal
∠AEC = ∠ECD by alternate angles
∴ ∠AEG + ∠AEC + ∠CED = 180° Sum of angles on a straight line
∠CDE = ∠CFD + ∠FCD
∠CDE + ∠CDF = 180° (linear pair angles)
∴ ∠AEG + ∠CDF = 180° by transitive property
∠AEC + ∠CED = ∠CDF by transitive property
∠AEC = ∠ECD = ∠CED = ∠CFD = r
∴ ∠CFD + ∠CFD = ∠CDF
2·r = ∠CDF
∠CDE = ∠FCD + ∠CFD
∠FCD = ∠CDF = 2·r
∴ ∠CDE = 2·r + r = 3·r
∠CDE = 3·r
The angles of the pentagon are;
∠CDE + ∠DCB + ∠ABC + ∠EAB + ∠AED = 540° sum of angles in a pentagon
∠DCB = 180° - ∠FCD
∠DCB = 180° - 2·r
∠ABC = ∠CDE = 3·r
∠EAB = ∠CEH corresponding angles
∠CEH = 180 - ∠AEC
∴ ∠EAB = 180° - ∠AEC
∴ ∠EAB = 180° - r
∠AED = ∠AEC + ∠CED = r + r = 2·r
∠AED = 2·r
Therefore, we have;
3·r + 180° - 2·r + 3·r + 180° - r + 2·r = 540°
5·r + 360° = 540°
r = (540° - 360°)/5 = 36°
r = 36°
∠CDE = 3 × 36° = 108°
∠DCB = 180° - 2×36° = 108°
∠ABC = 3 × 36° = 108°
∠EAB = 180° - 36° = 144°
∠AED = 2 × 36° = 72°
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