Question

Let ABCD be a cyclic quadrilateral satisfying AB = AD and AB + BC = CD.
Determine ∠CDA.

Answer 1

Hey there!

Given,

$\text{ABCD is a cyclic quadrilateral.}$

$AB = AD \ and \ AB+ BC = CD$

To find : ∠CDA

$\text{Suppose the point E on the segment CD such that DE = AD. }$

Refer to the image attached.

Then CE = CD − AD =  CD − AB = BC, and hence the triangle CEB is isosceles.

Now, since AB = AD then ∠BCA = ∠ACD.

This shows that CA is the bisector of ∠BCD = ∠BCE.

In an isosceles triangle, the bisector of the apex angle is also the  perpendicular bisector of the base.

Hence A is on the perpendicular bisector of BE,

and AE = AB = AD = DE.

This shows that triangle AED is equilateral, and thus  ∠CDA = 60°.

Good Studies!

Answer 2

Answer:

Given,

\text{ABCD is a cyclic quadrilateral.}ABCD is a cyclic quadrilateral.

AB = AD \ and \ AB+ BC = CDAB=AD and AB+BC=CD

To find : ∠CDA

\text{Suppose the point E on the segment CD such that DE = AD. }Suppose the point E on the segment CD such that DE = AD. 

Refer to the image attached.

Then CE = CD − AD =  CD − AB = BC, and hence the triangle CEB is isosceles.

Now, since AB = AD then ∠BCA = ∠ACD.

This shows that CA is the bisector of ∠BCD = ∠BCE.

In an isosceles triangle, the bisector of the apex angle is also the  perpendicular bisector of the base.

Hence A is on the perpendicular bisector of BE,

and AE = AB = AD = DE.

This shows that triangle AED is equilateral, and thus  ∠CDA = 60°.