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In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4
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Updated on: 12 Dec 2014, 07:43
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In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 40 degrees. What is the value of angle DAB? (1) AD is angular bisector of angle A. (2) AB x CD = AC x BD
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Originally posted by raj44 on 12 Dec 2014, 07:10.
Last edited by Bunuel on 12 Dec 2014, 07:43, edited 1 time in total.
Renamed the topic and edited the question.



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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4
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12 Dec 2014, 15:03
Can anybody explain why the answer is E?



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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4
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13 Dec 2014, 00:33
viktorija wrote: Can anybody explain why the answer is E? The answer is E because the question stem doesn't specify which 2 angles are equal, angle A or angle C could both be equal to 40. Also, the 2 statements mean the same AD as angle bisector itself means ab x bd = ac x dc



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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4
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17 Dec 2014, 06:55
raj44 wrote: viktorija wrote: Can anybody explain why the answer is E? The answer is E because the question stem doesn't specify which 2 angles are equal, angle A or angle C could both be equal to 40. Also, the 2 statements mean the same AD as angle bisector itself means ab x bd = ac x dc Unable to understand how ab x bd = ac x dc means AD is angle bisector? ab ac  =  bd dc means ABD and ACD are simliar triangles , not same triangles. Sides are in proportion, does it mean angles are same? Kindly clarify.



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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4
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17 Dec 2014, 08:59
anupamadw wrote: raj44 wrote: viktorija wrote: Can anybody explain why the answer is E? The answer is E because the question stem doesn't specify which 2 angles are equal, angle A or angle C could both be equal to 40. Also, the 2 statements mean the same AD as angle bisector itself means ab x bd = ac x dc Unable to understand how ab x bd = ac x dc means AD is angle bisector? ab ac  =  bd dc means ABD and ACD are simliar triangles , not same triangles. Sides are in proportion, does it mean angles are same? Kindly clarify. It's the angle bisector theorem. Check Wiki here: http://en.wikipedia.org/wiki/Angle_bisector_theorem
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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4
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25 May 2015, 02:04
raj44 wrote: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 40 degrees. What is the value of angle DAB?
(1) AD is angular bisector of angle A. (2) AB x CD = AC x BD Bunuel, I have one query here. Does the angle bisector extends 90 degree angle at base?



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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4
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25 May 2015, 02:17



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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4
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25 May 2015, 03:06
I hope its clear now.i have attached figure for same
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File comment: Does angle bisector extends 90 degree on the hypotenuse
IMG_20150525_153013_AO_HDR.jpg [ 1.33 MiB  Viewed 4225 times ]



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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4
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25 May 2015, 03:12



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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4
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26 May 2015, 08:11
ssriva2 wrote: I hope its clear now.i have attached figure for same Hi ssriva2, You can also understand this property in the following manner. Refer the diagram below: In the diagram, ABC is an isosceles triangle with sides AB = AC. As angles opposite the equal sides are equal we have ∠B = ∠C = \(x\). Also AD is the angle bisector of ∠A, therefore ∠BAD = ∠CAD = \(\frac{y}{2}\). We can also see from the above figure that ∠BDA = ∠CDA = \(z\) (as ∠BDA = ∠CDA = \(180  x  \frac{y}{2}\)) We also know that in triangle ABC, \(2x + y = 180\). Similarly in triangle ABD we can write \(z + x + \frac{y}{2} = 180\) i.e. \(2z + 2x + y = 360\). Since \(2x + y = 180\), we have \(2z + 180 = 360\) i.e. \(z = 90\). However please note that the only the angle bisector of the nonequal angle of an isosceles triangle will be perpendicular to the opposite base i.e. the nonequal side. Also AD bisects base BC i.e. BD = DC. These properties will not apply to the equal angles and the equal sides. In case of an equilateral triangle, we know that \(x = 60\) and \(\frac{y}{2} = 30\), hence \(z = 90\). In an equilateral triangle since all angles and sides are equal, these properties would apply to any of the angles and sides. Hope it's clear Regards Harsh
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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4
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05 Nov 2017, 03:07
EgmatQuantExpert wrote: ssriva2 wrote: I hope its clear now.i have attached figure for same Hi ssriva2, You can also understand this property in the following manner. Refer the diagram below: In the diagram, ABC is an isosceles triangle with sides AB = AC. As angles opposite the equal sides are equal we have ∠B = ∠C = \(x\). Also AD is the angle bisector of ∠A, therefore ∠BAD = ∠CAD = \(\frac{y}{2}\). We can also see from the above figure that ∠BDA = ∠CDA = \(z\) (as ∠BDA = ∠CDA = \(180  x  \frac{y}{2}\)) We also know that in triangle ABC, \(2x + y = 180\). Similarly in triangle ABD we can write \(z + x + \frac{y}{2} = 180\) i.e. \(2z + 2x + y = 360\). Since \(2x + y = 180\), we have \(2z + 180 = 360\) i.e. \(z = 90\). However please note that the only the angle bisector of the nonequal angle of an isosceles triangle will be perpendicular to the opposite base i.e. the nonequal side. Also AD bisects base BC i.e. BD = DC. These properties will not apply to the equal angles and the equal sides. In case of an equilateral triangle, we know that \(x = 60\) and \(\frac{y}{2} = 30\), hence \(z = 90\). In an equilateral triangle since all angles and sides are equal, these properties would apply to any of the angles and sides. Hope it's clear Regards Harsh So based on this itself why is it E? why not D?




Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4 &nbs
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