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ABCD is a parallelogram - Leaving Cert Mathematics - Question 6B - 2012

Question 6B

ABCD-is-a-parallelogram-Leaving Cert Mathematics-Question 6B-2012.png

ABCD is a parallelogram. The points A, B and C lie on the circle which cuts [AD] at P. The line CP meets the line BA at Q. Prove that $|CD| = |CP|$.

Worked Solution & Example Answer:ABCD is a parallelogram - Leaving Cert Mathematics - Question 6B - 2012

Step 1

Prove that $|AB| = |CD|$ and $|BC| = |AD|$

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Answer

Since ABCD is a parallelogram, opposite sides are equal. Therefore, we have:

$|AB| = |CD|$ .
$|BC| = |AD|$ .

This holds true as per the definitions of a parallelogram.

Step 2

Use properties of cyclic quadrilaterals

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Answer

From the properties of cyclic quadrilaterals, we know that the opposite angles sum up to 180 degrees. Thus, we state:

$|ABC| + |CPA| = 180^{ ext{o}}$ .
$|DPC| + |CBA| = 180^{ ext{o}}$ .

This indicates that $|ABP|$ and $|CDP|$ are alternate segments.

Step 3

Find the length relationships

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Answer

We can equate the lengths using the isosceles triangle theorem. Since angles at $P$ are equal:

$|CD| = |CP|$ due to structure of triangle $CDP$ being isosceles or resulting from chords in equal arcs.

Step 4

Conclusion

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Answer

Thus we have shown that:

$|CD| = |CP|$ .

This completes the proof.

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