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# In triangle ABC above

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Intern
Joined: 16 May 2018
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Updated on: 31 May 2018, 21:24
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33% (02:05) correct 67% (02:03) wrong based on 57 sessions

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In triangle ABC above, if AD=BC ,what is the value of x?
(1) D is the midpoint of side AC.
(2) The angle ABC is a right angle.

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Originally posted by jfdelgado on 31 May 2018, 18:22.
Last edited by amanvermagmat on 31 May 2018, 21:24, edited 1 time in total.
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Joined: 28 Nov 2017
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Updated on: 01 Jun 2018, 01:27
In triangle ABC above, if AD=BC ,what is the value of x?
(1) D is the midpoint of side AC.
(2) The angle ABC is a right angle.

$$ADB\angle=180-2x$$ and $$BAD\angle=180-ABD\angle-ADB\angle=x$$
If $$AD=BC$$, then $$ACB\angle=BAC\angle=x$$.

(1) If $$D$$ is the midpoint of side $$AC$$, then triangles $$ABD$$ and $$BDC$$ are identical. So, $$ADB\angle=2x$$ and $$x+x+2x=180$$. $$x=45$$. Sufficient.

(2) If the angle $$ABC\angle$$ is a right angle, then $$DBC\angle=90-x$$ and $$(90-x)+2x+x=180$$. $$x=45$$. Sufficient.

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Tulkin.

Originally posted by Tulkin987 on 31 May 2018, 21:34.
Last edited by Tulkin987 on 01 Jun 2018, 01:27, edited 1 time in total.
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Re: In triangle ABC above  [#permalink]

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31 May 2018, 21:49
Tulkin987 wrote:
In triangle ABC above, if AD=BC ,what is the value of x?
(1) D is the midpoint of side AC.
(2) The angle ABC is a right angle.

$$ADB=180-2x$$ and $$BAD=180-ABD-ADB=x$$
If $$AD=BC$$, then $$ACB=BAC=x$$.

(1) If $$D$$ is the midpoint of side $$AC$$, then triangles $$ABD$$ and $$BDC$$ are identical. So, $$ADB=2x$$ and $$x+x+2x=180$$. $$x=45$$. Sufficient.

(2) If the angle $$ABC$$ is a right angle, then $$DBC=90-x$$ and $$(90-x)+2x+x=180$$. $$x=45$$. Sufficient.

Hello

I also agree that answer would be D.
But wouldnt x be equal to 30 degrees from each of the statements. (I am getting 30 degrees)

Can you please check the highlighted part. By the word 'identical', do you mean that both triangles would be congruent? Because a median just divides the triangle into two triangles of equal area, not necessarily identical.
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Re: In triangle ABC above  [#permalink]

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31 May 2018, 21:58
amanvermagmat wrote:
Tulkin987 wrote:
In triangle ABC above, if AD=BC ,what is the value of x?
(1) D is the midpoint of side AC.
(2) The angle ABC is a right angle.

$$ADB=180-2x$$ and $$BAD=180-ABD-ADB=x$$
If $$AD=BC$$, then $$ACB=BAC=x$$.

(1) If $$D$$ is the midpoint of side $$AC$$, then triangles $$ABD$$ and $$BDC$$ are identical. So, $$ADB=2x$$ and $$x+x+2x=180$$. $$x=45$$. Sufficient.

(2) If the angle $$ABC$$ is a right angle, then $$DBC=90-x$$ and $$(90-x)+2x+x=180$$. $$x=45$$. Sufficient.

Hello

I also agree that answer would be D.
But wouldnt x be equal to 30 degrees from each of the statements. (I am getting 30 degrees)

Can you please check the highlighted part. By the word 'identical', do you mean that both triangles would be congruent? Because a median just divides the triangle into two triangles of equal area, not necessarily identical.

Hi,

Yes, based on the info given in the first statement triangles $$ABD$$ and $$BDC$$ are congruent, because $$AB=BC$$, $$AD=DC$$ and they share $$BD$$.

And please check $$x+x+2x=180$$. This will result in $$x=45$$, but not $$x=30$$.

Hope this helps!
_________________

Kindest Regards!
Tulkin.

DS Forum Moderator
Joined: 21 Aug 2013
Posts: 1412
Location: India
Re: In triangle ABC above  [#permalink]

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31 May 2018, 22:06

Hello

I also agree that answer would be D.
But wouldnt x be equal to 30 degrees from each of the statements. (I am getting 30 degrees)

Can you please check the highlighted part. By the word 'identical', do you mean that both triangles would be congruent? Because a median just divides the triangle into two triangles of equal area, not necessarily identical.[/quote]

Hi,

Yes, based on the info given in the first statement triangles $$ABD$$ and $$BDC$$ are congruent, because $$AB=BC$$, $$AD=DC$$ and they share $$BD$$.

And please check $$x+x+2x=180$$. This will result in $$x=45$$, but not $$x=30$$.

Hope this helps![/quote]

Hello

Actually triangle ABD has its angles measures as x, x and 180-2x (that is because exterior angle BDC is 2x, which is the sum of angle ABD and angle BAD, and since angle ABD is given as x, angle BAD also must be x). And also because of this, AD = BD = BC.

BUT triangle BDC has its angles measures as 2x, 2x and 180-4x (that is because BD=BC so both angles would be 2x each, leaving the third angle DBC to be 180-4x).

Since the angles differ, the triangles cannot be congruent. Thats what I think.
Intern
Joined: 30 May 2018
Posts: 1
Re: In triangle ABC above  [#permalink]

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01 Jun 2018, 01:18
Tulkin987 wrote:
In triangle ABC above, if AD=BC ,what is the value of x?
(1) D is the midpoint of side AC.
(2) The angle ABC is a right angle.

$$ADB=180-2x$$ and $$BAD=180-ABD-ADB=x$$
If $$AD=BC$$, then $$ACB=BAC=x$$.

(1) If $$D$$ is the midpoint of side $$AC$$, then triangles $$ABD$$ and $$BDC$$ are identical. So, $$ADB=2x$$ and $$x+x+2x=180$$. $$x=45$$. Sufficient.

(2) If the angle $$ABC$$ is a right angle, then $$DBC=90-x$$ and $$(90-x)+2x+x=180$$. $$x=45$$. Sufficient.

I'm not able to understand how ACB=BAC.
Manager
Joined: 28 Nov 2017
Posts: 145
Location: Uzbekistan
Re: In triangle ABC above  [#permalink]

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01 Jun 2018, 01:22
Chanda15 wrote:
Tulkin987 wrote:
In triangle ABC above, if AD=BC ,what is the value of x?
(1) D is the midpoint of side AC.
(2) The angle ABC is a right angle.

$$ADB=180-2x$$ and $$BAD=180-ABD-ADB=x$$
If $$AD=BC$$, then $$ACB=BAC=x$$.

(1) If $$D$$ is the midpoint of side $$AC$$, then triangles $$ABD$$ and $$BDC$$ are identical. So, $$ADB=2x$$ and $$x+x+2x=180$$. $$x=45$$. Sufficient.

(2) If the angle $$ABC$$ is a right angle, then $$DBC=90-x$$ and $$(90-x)+2x+x=180$$. $$x=45$$. Sufficient.

I'm not able to understand how ACB=BAC.

ACB and BAC are angles. Will correct.
_________________

Kindest Regards!
Tulkin.

Senior Manager
Joined: 14 Feb 2018
Posts: 395
Re: In triangle ABC above  [#permalink]

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01 Jun 2018, 05:38
According to statement 1, triangle BDC becomes equilateral triangle, and x = 30 not 45. If two sides are equal, doesn't mean the third angle BDA also becomes 2x.

Statement 2 is also sufficient.

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Joined: 02 Oct 2017
Posts: 727
Re: In triangle ABC above  [#permalink]

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30 Jun 2018, 10:07
amanvermagmat

If possible provide solution

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Re: In triangle ABC above &nbs [#permalink] 30 Jun 2018, 10:07
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