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

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Math Expert
Joined: 02 Sep 2009
Posts: 58396
Triangle ABC has a right angle at B, AB = 1, and BC = 2. The bisector  [#permalink]

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29 Mar 2019, 00:54
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Difficulty:

45% (medium)

Question Stats:

58% (02:24) correct 42% (02:35) wrong based on 12 sessions

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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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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.

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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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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.

EgmatQuantExpert
upon solving we get BD= 2/√5+1
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Joined: 04 Jan 2015
Posts: 3078
Re: Triangle ABC has a right angle at B, AB = 1, and BC = 2. The bisector  [#permalink]

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29 Mar 2019, 02:39
Archit3110 wrote:

EgmatQuantExpert
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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