Understanding Cyclic Quadrilaterals and Equilateral Triangles
The problem involves a cyclic quadrilateral ABCD and an equilateral triangle ABC. We need to find the measure of angle CDA.
Properties Utilized
- Equilateral Triangle: All angles in an equilateral triangle measure $60^{\circ}$.
- Cyclic Quadrilateral: The sum of opposite angles in a cyclic quadrilateral is $180^{\circ}$.
Step-by-Step Solution
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Since triangle ABC is equilateral, all its angles are $60^{\circ}$. Therefore, $\angle ABC = 60^{\circ}$.
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ABCD is a cyclic quadrilateral. A property of cyclic quadrilaterals is that opposite angles sum to $180^{\circ}$. This means $\angle ABC + \angle CDA = 180^{\circ}$.
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Substitute the known value of $\angle ABC$ into the equation:
$ 60^{\circ} + \angle CDA = 180^{\circ} $
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Solve for $\angle CDA$:
$ \angle CDA = 180^{\circ} - 60^{\circ} $
$ \angle CDA = 120^{\circ} $
Thus, the angle CDA is $120^{\circ}$.
Similar Questions
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- In a rhombus ABCD, if $\angle ACB = 40^\circ, \text{then } \angle ADB = ?$
- P is the mid-point of side BC of a parallelogram ABCD such that $\angle BAP = \angle DAP$.
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Important Questions from Quadrilaterals
What is the value of AC 2– BD 2
What is the point of intersection of the diagonals?
What is the area of the parallelogram?
ABCD is a cyclic quadrilateral. Diagonals BD and AC intersect each other at E. If ∠BEC = 138° and ∠ECD = 35°, then what is the measure of ∠BAC?
A circle is inscribed in a quadrilateral ABCD, touching sides AB, BC CD and DA at P, Q, R and S, respectively. If AS = 6 cm, BC = 12 cm, and CR = 5 cm, then the length of AB (in cm) is:
