Question: Possible Right Angle

Comment on Possible Right Angle

Can we say that since AC = 12 and BC = 5, this is a special triangle, 5-12-13 and is a right angle triangle and therefore, sufficient?
gmat-admin's picture

Unfortunately, that approach is not valid. You're assuming that angle ACB is a right angle in order to prove that angle ACB is a right angle.

Here's what I mean. We know that AC = 12 and BC = 5. However, we cannot conclude that the length of side AB is 13, because we don't know for certain that triangle ABC is a right triangle (in fact, the goal of this question is to determine whether or not triangle ABC is a right triangle).

So, we cannot assume that triangle ABC is a right triangle and then use this assumption to conclude that triangle ABC is a right triangle.

well, we cannot assume that because the hypotenuse can vary, right?
gmat-admin's picture

That's right; the hypotenuse (AB) can vary, because ∠ACB can vary.

When evaluating the statments together, Can we say that Triangle ABC and BCD are similar or equal and with this angles BCA and BCD measure the same and their sum iqual 180 degrees? so they sould be 90 degrees each.
gmat-admin's picture

We can't say that the triangles are similar, and we can't say that they're equal. This is because we can't show that the two triangles have any angles in common.

Hi Brent!
If BC divides AC in 2 equal parts, can't it be the bisector and then ACB be 90°?
gmat-admin's picture

Hi Laura,

Since BC divides AC in 2 equal parts, BC COULD be the bisector and ∠ACB COULD be 90°. However, that need not be the case.

IF it were true that AB = BD (making ∆ABD an isosceles triangle), then ∠ACB would definitely be 90°.

However, since we don't know whether AB = BD, we can't say for certain that ∠ACB = 90°.

Does that help?


Yes, thank you!

Hi Brent,

I selected B using the following reasoning. Please help what I missed out on:

Since the AC=CD, then the angle opposite them must also be equal i.e. angle(ABC)=angle(DBC), hence when we move the point B in either direction, the angles do not remain equal.

Warm Regards,
gmat-admin's picture

Be careful. If AC = CD, we can't necessarily conclude that ∠ABC = ∠DBC.

Hi Brent,

Doesn't it follow the rule that the angles opposite the sides with the length have the same angle?

Warm Regards,
gmat-admin's picture

That rule applies to 2 (of the 3 sides) of a triangle. More here:
In this case, AC and CD are not two sides of the same triangle.

We can say that AC and CD COMBINE to make ONE side of triangle ABD, but this doesn't really help us.

The lock rule is a game-changer. It's just such a simple way to think about these problems.

Thank you for that Brent!
gmat-admin's picture

I'm glad you like it, Erik!

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