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Two circles meet at points
X and
Y. Line segment
[AXB] meets one circle at
A and the other at
B. Line segment [CYD] meets one circle at
C and the other at
D. Prove that
[AC] is parallel to
[BD].

Two circles meet at points $\mathrm{X}$ and $\mathrm{Y}$ . Line segment $[\mathrm{AXB}]$ meets one circle at $\mathrm{A}$ and the other at $\mathrm{B}$ . Line segment [CYD] meets one circle at $C$ and the other at $D$ . Prove that $[A C]$ is parallel to $[B D]$ .

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Transformations of linear functions

Full solution

Q. Two circles meet at points

\mathrm{X}

and

\mathrm{Y}

. Line segment

[\mathrm{AXB}]

meets one circle at

\mathrm{A}

and the other at

\mathrm{B}

. Line segment [CYD] meets one circle at

C

and the other at

D

. Prove that

[A C]

is parallel to

[B D]

.

Identify Given Information: Identify the given information and the theorem that can be applied. $\newline$ In this case, we are dealing with two circles intersecting at points $X$ and $Y$ . The line segments $[AXB]$ and $[CYD]$ intersect the circles at points $A$ , $B$ , $C$ , and $D$ respectively. To prove that $[AC]$ is parallel to $[BD]$ , we can use the concept of alternate segment theorem which states that the angle between the tangent and chord at the point of contact is equal to the angle in the alternate segment.
Apply Alternate Segment Theorem: Apply the alternate segment theorem to the angles formed at points $A$ and $B$ . Since $[AXB]$ is a line segment intersecting the circles at $A$ and $B$ , angle $AXB$ is subtended by arc $XY$ in one circle, and angle $AYB$ is subtended by arc $XY$ in the other circle. According to the alternate segment theorem, angle $AXB$ is equal to angle $B$ $0$ , and angle $AYB$ is equal to angle $B$ $2$ .
Apply Theorem to Angles: Apply the alternate segment theorem to the angles formed at points $C$ and $D$ . Similarly, since $[CYD]$ is a line segment intersecting the circles at $C$ and $D$ , angle $CYD$ is subtended by arc $XY$ in one circle, and angle $CXD$ is subtended by arc $XY$ in the other circle. According to the alternate segment theorem, angle $CYD$ is equal to angle $D$ $0$ , and angle $CXD$ is equal to angle $D$ $2$ .
Combine Results for Parallel Lines: Combine the results from the alternate segment theorem to show parallel lines. $\newline$ From the previous steps, we have $\angle ADB = \angle AXB$ and $\angle ACB = \angle AYB$ . Also, $\angle CAD = \angle CYD$ and $\angle CBD = \angle CXD$ . Since angles $AXB$ and $AYB$ are the same, and angles $CYD$ and $CXD$ are the same, it follows that $\angle ADB = \angle ACB$ and $\angle CAD = \angle CBD$ . This means that angles $\angle ACB = \angle AYB$ $0$ and $\angle ACB = \angle AYB$ $1$ are alternate interior angles, and angles $\angle ACB = \angle AYB$ $2$ and $\angle ACB = \angle AYB$ $3$ are alternate interior angles.
Conclude Parallel Lines: Conclude that $[AC]$ is parallel to $[BD]$ . If alternate interior angles are equal, then the lines are parallel by the converse of the alternate interior angle theorem. Therefore, since angle $ADB = ACB$ and angle $CAD = CBD$ , line segment $[AC]$ is parallel to line segment $[BD]$ .

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