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Triangle ABC has a right angle at B, AB = 1, and BC = 2. The bisector

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Triangle ABC has a right angle at B, AB = 1, and BC = 2. The bisector  [#permalink]

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New post 29 Mar 2019, 00:54
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Question Stats:

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Triangle ABC has a right angle at B, AB = 1, and BC = 2. The bisector of angle BAC meets BC at D. What is BD ?


(A) \(\frac{\sqrt{3} - 1}{2}\)

(B) \(\frac{\sqrt{5} - 1}{2}\)

(C) \(\frac{\sqrt{5} + 1}{2}\)

(D) \(\frac{\sqrt{6} + \sqrt{2}}{2}\)

(E) \(2\sqrt{3} - 1\)

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Re: Triangle ABC has a right angle at B, AB = 1, and BC = 2. The bisector  [#permalink]

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New post 29 Mar 2019, 01:07

Solution


Given:
    • Triangle ABC is right-angled at B.
    • AB = 1 and BC = 2.
    • The bisector of angle BAC meets BC at D.

To find:
    • The length of BD.

Approach and Working:
As triangle ABC is right-angled at B, we can say
    • \(AC^2 = AB^2 + BC^2 = 1 + 4 = 5\)
    Or, AC = √5

Now, as per angle-bisector theorem,
    • \(\frac{AB}{AC} = \frac{BD}{DC}\)
    Or, \(\frac{1}{√5} = \frac{BD}{2 – BD}\)
    Simplifying, BD = \(\frac{√5 – 1}{2}\)

Hence, the correct answer is option B.

Answer: B

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Re: Triangle ABC has a right angle at B, AB = 1, and BC = 2. The bisector  [#permalink]

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New post 29 Mar 2019, 02:22
EgmatQuantExpert wrote:

Solution


Given:
    • Triangle ABC is right-angled at B.
    • AB = 1 and BC = 2.
    • The bisector of angle BAC meets BC at D.

To find:
    • The length of BD.

Approach and Working:
As triangle ABC is right-angled at B, we can say
    • \(AC^2 = AB^2 + BC^2 = 1 + 4 = 5\)
    Or, AC = √5

Now, as per angle-bisector theorem,
    • \(\frac{AB}{AC} = \frac{BD}{DC}\)
    Or, \(\frac{1}{√5} = \frac{BD}{2 – BD}\)
    Simplifying, BD = \(\frac{√5 – 1}{2}\)


Hence, the correct answer is option B.

Answer: B

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EgmatQuantExpert
please explain the highlighted part
upon solving we get BD= 2/√5+1
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Re: Triangle ABC has a right angle at B, AB = 1, and BC = 2. The bisector  [#permalink]

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New post 29 Mar 2019, 02:39
Archit3110 wrote:

EgmatQuantExpert
please explain the highlighted part
upon solving we get BD= 2/√5+1


Hey Archit3110,
following is the elaborate solution:
    • \(\frac{1}{√5} = \frac{BD}{2 – BD}\)
    Or, 2 – BD = √5BD
    Or, BD (√5 + 1) = 2
    Or, \(BD = \frac{2}{(√5 + 1)}\)
    Or, \(BD = \frac{2(√5 – 1)}{4} = \frac{√5 – 1}{2}\)

Hope it helps. :)
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Re: Triangle ABC has a right angle at B, AB = 1, and BC = 2. The bisector   [#permalink] 29 Mar 2019, 02:39
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Triangle ABC has a right angle at B, AB = 1, and BC = 2. The bisector

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