Question Video: Using Properties of Parallel Lines and Transversals to Find Congruent Angles | Nagwa Question Video: Using Properties of Parallel Lines and Transversals to Find Congruent Angles | Nagwa

Question Video: Using Properties of Parallel Lines and Transversals to Find Congruent Angles Mathematics • First Year of Preparatory School

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Draw △𝐴𝐵𝐶 where 𝐴𝐵 = 3 cm, 𝐵𝐶 = 4 cm, and 𝐴𝐶 = 5 cm. Point 𝐷 lies on the ray 𝐵𝐶 such that 𝐷 is not on line segment 𝐵𝐶. Draw ∠𝐷𝐶𝐸 congruent to ∠𝐶𝐵𝐴, where 𝐸 is on the upper side of line segment 𝐵𝐶. Which of the following is true? [A] 𝑚∠𝐴 = 𝑚∠𝐸𝐶𝐴 [B] 𝑚∠𝐴 = 𝑚∠𝐵𝐶𝐴 [C] 𝑚∠𝐴 = 𝑚∠𝐸𝐶𝐷 [D] 𝑚∠𝐵 = 𝑚∠𝐸𝐶𝐴

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Video Transcript

Draw triangle 𝐴𝐵𝐶 where 𝐴𝐵 equals three centimeters, 𝐵𝐶 equals four centimeters, and 𝐴𝐶 equals five centimeters. Point 𝐷 lies on the ray from 𝐵 through 𝐶 such that 𝐷 is not on line segment 𝐵𝐶. Draw angle 𝐷𝐶𝐸 congruent to angle 𝐶𝐵𝐴, where 𝐸 is on the upper side of line segment 𝐵𝐶. Which of the following is true? Option (A) the measure of angle 𝐴 equals the measure of angle 𝐸𝐶𝐴. Option (B) the measure of angle 𝐴 is equal to the measure of angle 𝐵𝐶𝐴. Option (C), the measure of angle 𝐴 equals the measure of angle 𝐸𝐶𝐷. Or is it option (D) the measure of angle 𝐵 is equal to the measure of angle 𝐸𝐶𝐴?

In this question, we’re given a construction and we need to determine which of five given options is correct in the construction. We need to start by sketching triangle 𝐴𝐵𝐶 by using the given side lengths of the triangle. We could do this by noting that the lengths of three, four, and five centimeters make a Pythagorean triple. This is enough to know that it is a right triangle with a hypotenuse of five centimeters by using the side-side-side congruency criterion.

However, it is not necessary to notice this to construct this triangle. In general, we can construct a triangle from its lengths by using circles. We start by drawing one of its sides. Let’s say 𝐴𝐵, which has a length of three centimeters. We can then trace a circle of radius four centimeters centered at 𝐵 and a circle of radius five centimeters centered at point 𝐴. Either point of intersection between these circles can be our point 𝐶. We choose 𝐶 as shown. We note that triangle 𝐴𝐵𝐶 has the desired lengths.

We now want to extend the line segment 𝐵𝐶 to be the ray from 𝐵 through 𝐶 so that we can find the point 𝐷 on this ray to make the angle 𝐷𝐶𝐸 congruent to the angle 𝐶𝐵𝐴. To do this, let’s start by highlighting the angle 𝐶𝐵𝐴 that we want to be congruent to the angle 𝐷𝐶𝐸. We know that this is a right angle since this is a Pythagorean triple. However, it is not necessary to use this to answer the question.

Instead, we can duplicate this angle of vertex 𝐶 by using our construction for duplicating an angle at a point. To duplicate this angle, we start by tracing a circle at 𝐵 with radius less than three centimeters. We call the points of intersection with the sides of the angle 𝐴 prime and 𝐶 prime. It is worth noting that it is easier to choose a radius below 1.5 centimeters.

Next, we trace a circle of the same radius centered at 𝐶. We label the point of intersection between the ray from 𝐵 through 𝐶 and the circle that is above 𝐶 as point 𝐷. Now, we can duplicate the angle shown at 𝐶 by tracing a circle of radius 𝐴 prime 𝐶 prime centered at 𝐷 and labeling the point of intersection between the circles 𝐸 as shown. We then know that angle 𝐷𝐶𝐸 is congruent to angle 𝐶𝐵𝐴. We can also see that line 𝐵𝐶 is a transversal of the lines 𝐶𝐸 and 𝐴𝐵. And this transversal makes the same angle with both lines. We can recall that this means that the two lines must be parallel.

We can then note that the line between 𝐴 and 𝐶 is also a transversal between these parallel lines. So the alternating angles it makes with each line must be congruent. This gives us that angle 𝐸𝐶𝐴 must have the same measure as angle 𝐴, which we can see is option (A).

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