Proof for salary of cyclic quadrilateral theorem
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Answer:there's your answer
Step-by-step explanation:
This article states that: "Another necessary and sufficient condition for a convex quadrilateral ABCD to be cyclic is that an angle between a side and a diagonal is equal to the angle between the opposite side and the other diagonal.[3] That is, for example,
∠ACB=∠ADB
"
I haven't found a proof for this online, can somebody proof this or find one?
Also, does this mean that if you draw in the diagonals, that of the 4 triangles you create the opposite ones are similar?
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Aug 19 '13 at 19:22
Prankster
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See the Wikipedia's inscribed angle article. – dtldarek Aug 19 '13 at 20:15
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This is based on the Inscribed Angle Theorem, which says that an inscribed angle measures half the angle of the arc that it subtends on the circle. The following proof is fairly simple.
Let D be the center of the circle. Note that ∠DAC=∠DCA and that
∠DAC+∠DCA+∠ADC=π=∠ADC+∠CDE(1)
enter image description here
Subtracting ∠ADC from both sides of (1) and remembering that ∠DAC=∠DCA, we have
∠CDE=2∠DAC
Similarly,
∠BDE=2∠DAB
Taking a sum or difference, we get that
∠CDB=2∠CAB
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In your diagram, this means that ∠ACB is half the arc from A to B. The same is true of ∠ADB.
To subtend the same arc of a circle, the vertices of the angles must be on the same side of the chord between the ends of the arc. To prove the converse, we need to assume that the points considered be on the same side of the chord.
Suppose we have a point C on a circle with a given ∠ACB. Suppose that ∠ADB is the same angle, but D is not on the circle.
enter image description here
Find the point E at the intersection of BD¯¯¯¯¯¯¯¯ and the circle. We know by the Inscribed Angle Theorem, that ∠AEB=∠ACB. However, if D is outside the circle, ∠AEB>∠ADB, and if D is inside the circle, ∠AEB<∠ADB. Therefore, D must be on the circle.
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Thus, if two points are on the same side of the chord, and the chord subtends the same angle at both, then they are on the same circle with that chord.