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ABCD is a parallelogram. ∠C = 110°. E is a point on DC so that AD = ED. Find ∠AEC. - Mathematics

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Question

ABCD is a parallelogram. ∠C = 110°. E is a point on DC so that AD = ED. Find ∠AEC.

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Solution

Given:

  • ABCD is a parallelogram.
  • ∠C = 110°.
  • E is a point on DC such that AD = ED.

Stepwise calculation:

  1. Since ABCD is a parallelogram, opposite angles are equal. So ∠A = ∠C = 110°.
  2. Adjacent angles in a parallelogram sum to 180°, so ∠D = 180° – 110° = 70°.
  3. Since AD = ED, triangle ADE is isosceles with AD = ED.
  4. ∠ADE = ∠DEA because of the isosceles triangle property.
  5. Angle ∠ADC whole angle at D = ∠ADE + ∠EDC = 70° and since ∠EDC = ∠C = 110°, point E lies on DC such that DE = AD.
  6. To find ∠AEC, we use the exterior angle theorem in triangle AED.
    • ∠AEC = 180° – ∠DEA since E, A, C are points in a polygon.
  7. From the isosceles triangle ADE, calculate ∠ADE and ∠DEA
    = `(180^circ - 70^circ)/2`
    = 55°
  8. Therefore, ∠AEC = 180° – 55° = 125°.

∠AEC = 125°.

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Chapter 12: Rectilinear Figures (Theorems on Parallelograms and Construction of Polygons) - MISCELLANEOUS EXERCISE [Page 152]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 12 Rectilinear Figures (Theorems on Parallelograms and Construction of Polygons)
MISCELLANEOUS EXERCISE | Q I. 1. | Page 152
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