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

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In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4 [#permalink]

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Dec 2014, 06:43, 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 [#permalink]

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New post 12 Dec 2014, 14:03
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 [#permalink]

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New post 12 Dec 2014, 23:33
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 [#permalink]

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New post 17 Dec 2014, 05:55
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 [#permalink]

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New post 17 Dec 2014, 07:59
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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 [#permalink]

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New post 25 May 2015, 01:04
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 [#permalink]

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New post 25 May 2015, 01:17
ssriva2 wrote:
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?


Sorry, I don't understand what you mean...
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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4 [#permalink]

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New post 25 May 2015, 02:06
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
IMG_20150525_153013_AO_HDR.jpg [ 1.33 MiB | Viewed 2359 times ]

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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4 [#permalink]

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New post 25 May 2015, 02:12

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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4 [#permalink]

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New post 26 May 2015, 07:11
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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:

Image

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 non-equal angle of an isosceles triangle will be perpendicular to the opposite base i.e. the non-equal 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 [#permalink]

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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4 [#permalink]

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New post 05 Nov 2017, 02:07
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:

Image

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 non-equal angle of an isosceles triangle will be perpendicular to the opposite base i.e. the non-equal 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?

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Re: In an Isosceles Triangle ABC, point D lies on segment BC. Angle B is 4   [#permalink] 05 Nov 2017, 02:07
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