Math Problem Statement

In Figure 1, ABCD is a circle. E is a point lying on AC such that BC = CE. It is given that AB = AD, ∠ADB = 58° and ∠CBD = 25°. Find ∠BDC and ∠ABE.

Solution

To solve for BDC\angle BDC and ABE\angle ABE, let's analyze the information given in the diagram and the properties of the circle and the angles.

Step-by-Step Solution

  1. Understanding Given Information:

    • ABCDABCD is a circle.
    • EE is a point on line ACAC such that BC=CEBC = CE.
    • AB=ADAB = AD.
    • ADB=58\angle ADB = 58^\circ and CBD=25\angle CBD = 25^\circ.
  2. Finding BDC\angle BDC:

    • Since ABCDABCD is a circle, and points BB, DD, and CC are on the circumference, we can use the property of a cyclic quadrilateral.
    • In a cyclic quadrilateral, the opposite angles add up to 180180^\circ.
    • Thus, ADB+BDC=180\angle ADB + \angle BDC = 180^\circ.
    • Given ADB=58\angle ADB = 58^\circ, we have: BDC=18058=122\angle BDC = 180^\circ - 58^\circ = 122^\circ
  3. Finding ABE\angle ABE:

    • Since BC=CEBC = CE, triangle BCEBCE is isosceles.
    • Therefore, CBE=CEB\angle CBE = \angle CEB.
    • We are given CBD=25\angle CBD = 25^\circ, which means CBE=25\angle CBE = 25^\circ (as CBE\angle CBE and CBD\angle CBD are the same in this isosceles setup).
    • ABE\angle ABE is the external angle for BCE\triangle BCE at point BB, so: ABE=CBD+CBE=25+25=50\angle ABE = \angle CBD + \angle CBE = 25^\circ + 25^\circ = 50^\circ

Final Answers

  • BDC=122\angle BDC = 122^\circ
  • ABE=50\angle ABE = 50^\circ

Would you like a detailed explanation of the concepts used, or do you have any questions?

Here are some related questions that expand on this topic:

  1. How does the property of opposite angles in a cyclic quadrilateral help in solving circle geometry problems?
  2. What are the properties of an isosceles triangle that are useful in circle geometry?
  3. How would you determine angles in a cyclic quadrilateral if no values were given initially?
  4. Can external angles of polygons within circles help determine unknown internal angles?
  5. What other properties of cyclic quadrilaterals can be applied in solving complex geometric problems?

Tip: In a circle, knowing a few properties of cyclic quadrilaterals can greatly simplify finding unknown angles. Look for opposite angles and isosceles triangles to leverage symmetry in your solutions.

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Math Problem Analysis

Mathematical Concepts

Circle Geometry
Cyclic Quadrilateral
Isosceles Triangle
External Angles

Formulas

∠ADB + ∠BDC = 180° (opposite angles in a cyclic quadrilateral)
External angle of triangle = sum of two opposite interior angles

Theorems

Cyclic Quadrilateral Theorem
Isosceles Triangle Properties
External Angle Theorem

Suitable Grade Level

Grades 10-12