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705-805 (Hard)|   Geometry|      
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chetan2u
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Common Mistakes in Geometry Questions - Exercise Question #2 06 May 2020, 05:01
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elPatron434
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Hi Bunuel, chetan2u, VeritasKarishma
In STMT 2,
It is provided that AB=BD=CB,
Which means that Angle CAB = Angle ACB = Angle BCD = Angle CDB = x.

So would it be incorrect to assume that the triangles ACB and CBD are congruent(Angle CAB = Angle CDB and Angle ACB = Angle BCD ) wih BC being the common side. For this to be possible Angle CBA needs to be a right angle and Angle CDB hence becomes 45

Even solving it geometrically,
We would be arriving at Angle CBA = 180 -2x(Angle sum property)
And Angle CBD to be 2x

In Triangle BCD, Angle CDB and Angle BCD are equal to x
so 4x =180
x= 45 and 180-2x =90

Consequently, I feel that the Answer should be D. Could you please point out where I am going wrong?

Thank You
Show more


Hi

You are wrong in interpreting AB=BD=BC as all four angles opposite are equal.
You can correlate angles in two different triangles by this. Yes both the triangles ABC and BCD are isosceles. That’s it.

Imagine there is a straight piece of wood AD and BC is half of it.
Place B as a hinge in center of AD. Now you can rotate other end that is C anywhere from A to D. At all points where C moves from A to D the three sides AB, BC, and CD are equal but the angles will change in each triangle. It is only when that CB is perpendicular on AD that all four angles will be equal.
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Common Mistakes in Geometry Questions - Exercise Question #2 06 May 2020, 05:08
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chetan2u
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Hi Bunuel, chetan2u, VeritasKarishma
In STMT 2,
It is provided that AB=BD=CB,
Which means that Angle CAB = Angle ACB = Angle BCD = Angle CDB = x.

So would it be incorrect to assume that the triangles ACB and CBD are congruent(Angle CAB = Angle CDB and Angle ACB = Angle BCD ) wih BC being the common side. For this to be possible Angle CBA needs to be a right angle and Angle CDB hence becomes 45

Even solving it geometrically,
We would be arriving at Angle CBA = 180 -2x(Angle sum property)
And Angle CBD to be 2x

In Triangle BCD, Angle CDB and Angle BCD are equal to x
so 4x =180
x= 45 and 180-2x =90

Consequently, I feel that the Answer should be D. Could you please point out where I am going wrong?

Thank You


Hi

You are wrong in interpreting AB=BD=BC as all four angles opposite are equal.
You can correlate angles in two different triangles by this. Yes both the triangles ABC and BCD are isosceles. That’s it.

Imagine there is a straight piece of wood AD and BC is half of it.
Place B as a hinge in center of AD. Now you can rotate other end that is C anywhere from A to D. At all points where C moves from A to D the three sides AB, BC, and CD are equal but the angles will change in each triangle. It is only when that CB is perpendicular on AD that all four angles will be equal.
Show more


Thank You VeritasKarishma,
That clarifies things. I missed considering this. Like you said we just know the lengths of the 2 sides and the third side can range from (0,2*length of side).
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Common Mistakes in Geometry Questions - Exercise Question #2 06 May 2020, 05:38
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elPatron434
chetan2u



Hi

You are wrong in interpreting AB=BD=BC as all four angles opposite are equal.
You can correlate angles in two different triangles by this. Yes both the triangles ABC and BCD are isosceles. That’s it.

Imagine there is a straight piece of wood AD and BC is half of it.
Place B as a hinge in center of AD. Now you can rotate other end that is C anywhere from A to D. At all points where C moves from A to D the three sides AB, BC, and CD are equal but the angles will change in each triangle. It is only when that CB is perpendicular on AD that all four angles will be equal.


Thank You VeritasKarishma,
That clarifies things. I missed considering this. Like you said we just know the lengths of the 2 sides and the third side can range from (0,2*length of side).
Show more


Hi

1) you are wrong on two issues..
1) it is chetan :)
2) No, all three sides are equal but the side BC can be placed in any way. so angles will change. I have just added two case. you can have thousands cases.
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elPatron434
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Common Mistakes in Geometry Questions - Exercise Question #2 06 May 2020, 05:42
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chetan2u
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Hi

You are wrong in interpreting AB=BD=BC as all four angles opposite are equal.
You can correlate angles in two different triangles by this. Yes both the triangles ABC and BCD are isosceles. That’s it.

Imagine there is a straight piece of wood AD and BC is half of it.
Place B as a hinge in center of AD. Now you can rotate other end that is C anywhere from A to D. At all points where C moves from A to D the three sides AB, BC, and CD are equal but the angles will change in each triangle. It is only when that CB is perpendicular on AD that all four angles will be equal.


Thank You VeritasKarishma,
That clarifies things. I missed considering this. Like you said we just know the lengths of the 2 sides and the third side can range from (0,2*length of side).


Hi

1) you are wrong on two issues..
1) it is chetan :)
2) No, all three sides are equal but the side BC can be placed in any way. so angles will change. I have just added two case. you can have thousands cases.
Show more

My sincerest apologies chetan2u, guess I wasnt looking straight.

Thank you for the clarification on the sides of the triangle. I understand the conceot much better now!
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Common Mistakes in Geometry Questions - Exercise Question #2 08 May 2020, 02:54
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elPatron434
Bunuel
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Having a very hard time understanding any of the explanations of why statement 2 is insufficient. The question seems to imply that angle ACD is 90 degrees. If AB = BC = BD, doesn't that mean that angle ACB + BCD = 90?

Yes, angle ACD is 90 degrees and ACB + BCD = 90 degrees but the point is that we don't know whether CB is perpendicular to AD. Not knowing that, how are you getting that (2) is sufficient?

Hi Bunuel,

Its given,AB=BC=BD
So here, it satisfies this equation: BC^2=AB.BD
This implies, CB is perpendicular to AD.

Can you please tell whether this understanding is flawed?

Useful property: The median on the hypotenuse of a right triangle always equals to one-half the hypotenuse.

AB = BC = BD means that CB is the median (B is the midpoint of AD). Because of the property above, AB = BC = BD is true for any right angled triangle with median CB. So, CB may or may not be perpendicular to AD.

Hi Bunuel, chetan2u, VeritasKarishma
In STMT 2,
It is provided that AB=BD=CB,
Which means that Angle CAB = Angle ACB = Angle BCD = Angle CDB = x.

So would it be incorrect to assume that the triangles ACB and CBD are congruent(Angle CAB = Angle CDB and Angle ACB = Angle BCD ) wih BC being the common side. For this to be possible Angle CBA needs to be a right angle and Angle CDB hence becomes 45

Even solving it geometrically,
We would be arriving at Angle CBA = 180 -2x(Angle sum property)
And Angle CBD to be 2x

In Triangle BCD, Angle CDB and Angle BCD are equal to x
so 4x =180
x= 45 and 180-2x =90

Consequently, I feel that the Answer should be D. Could you please point out where I am going wrong?

Thank You
Show more

Here is the source of all trouble: Angle ABC LOOKS 90 degrees but may not be. Your analysis will change the moment I give you some other figures that could represent the given data.
Check this post by fskilnik above: https://gmatclub.com/forum/common-mista ... l#p2160694
Here Fabio gives you some different ways in which you can draw this triangle. Now would you say that statement 2 is enough?
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Common Mistakes in Geometry Questions - Exercise Question #2 28 May 2021, 11:49
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Hi ,

The answer is A because when CB bisects AD, we are not sure id CB makes a 90 degree or not and if B is not 90 degree, then we cannot be sure if angle CBD and angle BDC are 45 degree according to isosceles property. Hence angle BDC can be 60,30 or anything as angle B is not 90
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Common Mistakes in Geometry Questions - Exercise Question #2 12 Oct 2022, 05:29
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