Is angle CBD a right angle?
1). AE is parallel to BD
2). BC=BD
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Mike@Magoosh
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Hi, there. I'm happy to help with this. 
From the diagram itself, we know that the circle is inscribed in the triangle, which is to say, the three sides of the triangle are tangent to the circle. Side BC is tangent to the circle at point E, so angle AEC must be 90 degrees. This is because a radius is always perpendicular to a tangent line at the point of tangency.
We want to know: Is angle CBD a right angle?
Statement #1: AE is parallel to BD
Well, AE is perpendicular to BC, and if AE is parallel to BD, that means BD must also be perpendicular to BC. Two parallel lines have to be perpendicular to the same line. If BD is perpendicular to BC, that means the angle where they meet, angle CBD, is a right angle. This statement is sufficient.
Statement #2: BC=BD
Well, that would tell us that triangle BCD is an isosceles triangle, and angle BCD = angle BDC. Angle CBD would be the vertex angle of an isosceles triangle. The vertex angle of an isosceles triangle can be acute, right, or obtuse, and in any of those cases, it's possible to inscribe a circle into the triangle. This statement gives us absolutely no information about whether angle CBD is right or not. This statement is insufficient.
Answer = A.
Please let me know if you have any questions.
Mike
From the diagram itself, we know that the circle is inscribed in the triangle, which is to say, the three sides of the triangle are tangent to the circle. Side BC is tangent to the circle at point E, so angle AEC must be 90 degrees. This is because a radius is always perpendicular to a tangent line at the point of tangency.
We want to know: Is angle CBD a right angle?
Statement #1: AE is parallel to BD
Well, AE is perpendicular to BC, and if AE is parallel to BD, that means BD must also be perpendicular to BC. Two parallel lines have to be perpendicular to the same line. If BD is perpendicular to BC, that means the angle where they meet, angle CBD, is a right angle. This statement is sufficient.
Statement #2: BC=BD
Well, that would tell us that triangle BCD is an isosceles triangle, and angle BCD = angle BDC. Angle CBD would be the vertex angle of an isosceles triangle. The vertex angle of an isosceles triangle can be acute, right, or obtuse, and in any of those cases, it's possible to inscribe a circle into the triangle. This statement gives us absolutely no information about whether angle CBD is right or not. This statement is insufficient.
Answer = A.
Please let me know if you have any questions.
Mike
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vishal.pathak
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Hello Mike,mikemcgarry wrote:Hi, there. I'm happy to help with this.
From the diagram itself, we know that the circle is inscribed in the triangle, which is to say, the three sides of the triangle are tangent to the circle. Side BC is tangent to the circle at point E, so angle AEC must be 90 degrees. This is because a radius is always perpendicular to a tangent line at the point of tangency.
We want to know: Is angle CBD a right angle?
Statement #1: AE is parallel to BD
Well, AE is perpendicular to BC, and if AE is parallel to BD, that means BD must also be perpendicular to BC. Two parallel lines have to be perpendicular to the same line. If BD is perpendicular to BC, that means the angle where they meet, angle CBD, is a right angle. This statement is sufficient.
Statement #2: BC=BD
Well, that would tell us that triangle BCD is an isosceles triangle, and angle BCD = angle BDC. Angle CBD would be the vertex angle of an isosceles triangle. The vertex angle of an isosceles triangle can be acute, right, or obtuse, and in any of those cases, it's possible to inscribe a circle into the triangle. This statement gives us absolutely no information about whether angle CBD is right or not. This statement is insufficient.
Answer = A.
Please let me know if you have any questions.
Mike
Please correct me
I think that tangents are always perpendicular to the radius. So AE is perpendicular to BC. Similarly, the radius touching BD (lets call it AF) will be perpendicular to BD
So AFBE will have opposite angles equal to 90 degree. The 2 radius will b equal. Again BE and BF will be equal because tangents to the same circle from the same point are equal.
So all this makes it obvious that AFBE is a square and I didnt even require any of the statements for this. I am making some blunder. Please point it out
Regards,
Vishal
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ArunangsuSahu
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Pretty Simple:
Statement 1:
AE is perpendicular to BC as AE is in-radius and BC is the tangent to the incircle
AE||BD ....Sufficient
Statement 2:
It just tells Isosceles not right-angled isosceles specifically...Insufficient
So (A)
Statement 1:
AE is perpendicular to BC as AE is in-radius and BC is the tangent to the incircle
AE||BD ....Sufficient
Statement 2:
It just tells Isosceles not right-angled isosceles specifically...Insufficient
So (A)
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Mike@Magoosh
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This is a reply to Vishal ---
First of all, you are correct that AE is perpendicular to BC, and if we constructed the radius on the other side, we would get another radius AF that would be perpendicular to BD. Perfectly true. AE would have to equal AF, because they are both radii, and BE would have to equal BF, because they are both tangent segments from the same point. All 100% completely true.
Here's the rub though --- the fact that angle AEF = angle AFB = 90 degree AND AE = AF AND BE = BF is not enough to guarantee that AFBE is a square. AFBE would technically be something called a "kite", and only if all four sides were congruent would it be both a rhombus & a square. I've included a diagram with two figures, one showing AFBE in a near square configuration, and another showing a kite AFBE that meets all of these conditions and is clearly not a square. Very tricky.
Because we don't know whether AFBE is a square, we don't know whether or not angle CBD is 90 degrees. That's why, unless we have statement #1, we cannot draw any conclusions about angle CBD.
Does that make sense? Please let me know if you have any questions about what I've said here, or any questions at all about this problem.
Mike
First of all, you are correct that AE is perpendicular to BC, and if we constructed the radius on the other side, we would get another radius AF that would be perpendicular to BD. Perfectly true. AE would have to equal AF, because they are both radii, and BE would have to equal BF, because they are both tangent segments from the same point. All 100% completely true.
Here's the rub though --- the fact that angle AEF = angle AFB = 90 degree AND AE = AF AND BE = BF is not enough to guarantee that AFBE is a square. AFBE would technically be something called a "kite", and only if all four sides were congruent would it be both a rhombus & a square. I've included a diagram with two figures, one showing AFBE in a near square configuration, and another showing a kite AFBE that meets all of these conditions and is clearly not a square. Very tricky.
Because we don't know whether AFBE is a square, we don't know whether or not angle CBD is 90 degrees. That's why, unless we have statement #1, we cannot draw any conclusions about angle CBD.
Does that make sense? Please let me know if you have any questions about what I've said here, or any questions at all about this problem.
Mike
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