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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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Updated on: 03 Jun 2014, 01:07
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45% (medium)

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71% (01:26) correct 29% (01:47) wrong based on 156 sessions

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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°

Attachment:
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12.jpg [ 39.5 KiB | Viewed 2663 times ]
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 or else any one can correct me

Originally posted by awal_786@hotmail.com on 02 Jun 2014, 16:21.
Last edited by Bunuel on 03 Jun 2014, 01:07, edited 1 time in total.
Renamed the topic and edited the question.
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Joined: 18 Feb 2013
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Re: Found Mistake in 800Score CAT  [#permalink]

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02 Jun 2014, 17:03
2
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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02 Jun 2014, 21:08
3
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 or else any one can correct me

Check out the figure given below:
Attachment:

Ques3.jpg [ 12.8 KiB | Viewed 2803 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.

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

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30 Oct 2017, 11:06
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•)

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

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18 Dec 2018, 06:16
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).
Re: AD = BD = CD. If angle BAD = 50°, what is angle BCA?   [#permalink] 18 Dec 2018, 06:16
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# AD = BD = CD. If angle BAD = 50°, what is angle BCA?

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