Given the points B and C below, construct, without using a protractor or setsquare, a point A such that $|\angle ABC| = 60°$. Hence construct, on the same diagram above, and using a compass and straight edge only, an angle of 15°. In the diagram, $l_1, l_2, l_3, l_4$ are parallel lines that make intercepts of equal length on the transversal k. FG is parallel to k, and HG is parallel to ED. Prove that the triangles $\triangle CDE$ and $\triangle FGH$ are congruent.

Leaving Cert Mathematics Geometry: Given the points B and C below,

Given the points B and C below, construct, without using a protractor or setsquare, a point A such that $|\angle ABC| = 60°$ - Leaving Cert Mathematics - Question 6A - 2017

Question 6A

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Given the points B and C below, construct, without using a protractor or setsquare, a point A such that $|\angle ABC| = 60°$. Hence construct, on the same diagr... show full transcript

Worked Solution & Example Answer:Given the points B and C below, construct, without using a protractor or setsquare, a point A such that $|\angle ABC| = 60°$ - Leaving Cert Mathematics - Question 6A - 2017

Step 1

Construct a point A such that $|\angle ABC| = 60°$

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Answer

Start by placing points B and C on the paper with a ruler.
Use a compass and set its width to a convenient length. Label it as point A.
From point B, draw an arc above the line BC.
Without changing the compass width, set the compass point at A and draw an arc that intersects the first arc. Label the intersection point as D.
The angle (|\angle ABC|) is formed by line AB and line AC at point B. Adjust point A on the arc until (|\angle ABC| = 60°).
Draw lines AB and AC to complete the construction.

Step 2

Hence construct, on the same diagram above, and using a compass and straight edge only, an angle of 15°.

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Answer

Take the original 60° angle constructed.
Bisect the angle (|\angle ABC|) using a compass by making an arc across both rays. Label intersection points as E and F.
Now take the angle formed by AE and AB, which is 30°.
To get 15°, bisect the newly formed angle (30°) again using the same method. The intersection will form an angle of 15°.

Step 3

Prove that the triangles $\triangle CDE$ and $\triangle FGH$ are congruent.

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Answer

Notice that lines CD and GH are opposite sides of the parallelogram created by the parallel lines, implying (|CD| = |GH|).
Since FG is parallel to k, it makes matching intercepts on both lines, thus (|FG| = |ED|).
The corresponding angles (\angle CDE = \angle FGH) because they are alternate interior angles created by the parallel lines.
Therefore, by the Angle-Side-Angle (ASA) criterion of congruence, the triangles (\triangle CDE) and (\triangle FGH) are congruent.

Given the points B and C below, construct, without using a protractor or setsquare, a point A such that \(|\angle ABC| = 60°\) - Leaving Cert Mathematics - Question 6A - 2017

Worked Solution & Example Answer:Given the points B and C below, construct, without using a protractor or setsquare, a point A such that \(|\angle ABC| = 60°\) - Leaving Cert Mathematics - Question 6A - 2017

Construct a point A such that \(|\angle ABC| = 60°\)

Hence construct, on the same diagram above, and using a compass and straight edge only, an angle of 15°.

Prove that the triangles \(\triangle CDE\) and \(\triangle FGH\) are congruent.

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