Focus on triangle QAC. We're given that CQ = CA, which in turn means the angles opposite these sides (angles A and Q in that order) are congruent. This is one property of isosceles triangles. Let's make x equal to each of these angles.
With inscribed angle QAC = x, it doubles to 2x which is the measure of minor arc BC (inscribed angle theorem). Minor arc AB is also 2x, since we're told these two arcs have the same measure.
If minor arc BC = 2x, then cutting this in half leads to x again, which points to inscribed angle BDC equal to x. Notice how inscribed angles QAC and BDC subtend the same minor arc BC. Through similar logic, inscribed angle BDA is also x.
In short: angle BDC = angle BDA = x. This leads to angle ADC = 2x. We'll use this later.
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Let's go back to triangle QAC. The angles A and Q are both x each. The missing angle C is...
Q+A+C = 180
x+x+C = 180
2x+C = 180
C = 180-2x
Making angle QCA equal to 180-2x
Angle ACD is supplementary to this.
So,
angle ACD = 180-(angle QCA) = 180-(180-2x) = 2x. We'll use this later as well.
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The first section concluded with angle ADC = 2x, while the second section concluded with angle ACD = 2x
For triangle ACD, the bottom angles C and D are both 2x. This is sufficient evidence to show that triangle ACD is isosceles. Furthermore, it means AC = AD, which are the sides opposite angles D and C respectively.