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AD = BD = CD. If angle BAD = 50°, what is angle BCA?

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AD = BD = CD. If angle BAD = 50°, what is angle BCA?  [#permalink]

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New post Updated on: 03 Jun 2014, 00:07
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AD = BD = CD. If angle BAD = 50°, what is angle BCA?
(Note: figure not drawn to scale.)

A. 30°
B. 40°
C. 50°
D. 60°
E. 70°


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Answer given by software looks wrong to me, according to the property that, when we have equal sides there opposite angles should be equal. and if we solve this problem it should be 30, but software gives different answer, If I really pointed out mistake of 800Score flash CAT then I deserve few kudos :-D or else any one can correct me :)

Originally posted by awal_786@hotmail.com on 02 Jun 2014, 15:21.
Last edited by Bunuel on 03 Jun 2014, 00:07, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Found Mistake in 800Score CAT  [#permalink]

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New post 02 Jun 2014, 16:03
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I got B as the answer.

AD=DB so it is an isolate triangle. We now BAD is 50 so ABD is also 50, and ADB= 80.
If ADB=80 then BDC=100.
We also know BD=DC so it is another isolate triangle. 180-100=80, 80/2=40 for DBC and DCB.
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Re: Found Mistake in 800Score CAT  [#permalink]

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New post 02 Jun 2014, 20:08
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awal_786@hotmail.com wrote:
Answer given by software looks wrong to me, according to the property that, when we have equal sides there opposite angles should be equal. and if we solve this problem it should be 30, but software gives different answer, If I really pointed out mistake of 800Score flash CAT then I deserve few kudos :-D or else any one can correct me :)


Check out the figure given below:
Attachment:
Ques3.jpg
Ques3.jpg [ 12.8 KiB | Viewed 2356 times ]

Since AD = BD, opposite angles are 50 degrees each.
Sum of all angles in a triangle is 180 so leftover angle in ABD is 80 degrees.
BDC is then 100 degrees because it is a straight angle with ABD.

In triangle BDC, since BD = DC, both angles must be 40 degrees each.

Answer (B)
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AD = BD = CD. If angle BAD = 50°, what is angle BCA?  [#permalink]

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New post 30 Oct 2017, 10:06
Assuming that there's a circle subscribing the traingle ABC with center D and radious DA = DC= DB.

AC must be a diagonal hence, Angle ABC is a 90• and Traingle ABC is a right angled traingle. Now it's given that angle BAD is 50, so we can calculate Angle BCA (180-90-50 = 40•)

[size=80][b][i]Posted from my mobile device[/i][/b][/size]
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Re: AD = BD = CD. If angle BAD = 50°, what is angle BCA?  [#permalink]

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New post 18 Dec 2018, 05:16
The diagram shows that BD bisects AC into 2 and is actually equal to AD and DC.

This immediately shows that ABC is a right-angled triangle where BD = median to hypotenuse (AC) where median is essentially half the length of hypotenuse (AC). Therefore angle BCA is essentially 40 degrees (without using calculation of the angles of the smaller triangles).
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Re: AD = BD = CD. If angle BAD = 50°, what is angle BCA? &nbs [#permalink] 18 Dec 2018, 05:16
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AD = BD = CD. If angle BAD = 50°, what is angle BCA?

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