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# In the figure, if line CE bisects ∠ACB, then x =

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Math Expert
Joined: 02 Sep 2009
Posts: 58459
In the figure, if line CE bisects ∠ACB, then x =  [#permalink]

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15 Oct 2018, 07:57
00:00

Difficulty:

15% (low)

Question Stats:

93% (01:24) correct 7% (00:56) wrong based on 33 sessions

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In the figure, if line CE bisects ∠ACB, then x =

(A) 45

(B) 50

(C) 55

(D) 65

(E) 70

Attachment:

triangle.jpg [ 18.02 KiB | Viewed 409 times ]

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Re: In the figure, if line CE bisects ∠ACB, then x =  [#permalink]

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15 Oct 2018, 08:19
Bunuel wrote:

In the figure, if line CE bisects ∠ACB, then x =

(A) 45

(B) 50

(C) 55

(D) 65

(E) 70

Given, line CE bisects ∠ACB. Hence ∠ACE=∠ECB

In the right angled triangle EBC, ∠ECB=180-90-x=90-x=∠ACE

Now, BD is perpendicular to hypotenuse AC of right angled triangle ABC.
Therefore, the triangle ABC is divided into two parts, i.e., Triangle ADB and Triangle BDC.
Triangle ADB is similar to the Triangle BDC. Hence on the basis of similarity, corresponding angles are equal
So, ∠DCB=90-x+90-x=180-(90+40)=180-130=50
Or, 180-2x=50
Or, 2x=130
Or, x=65

Ans. (D)
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Re: In the figure, if line CE bisects ∠ACB, then x =  [#permalink]

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15 Oct 2018, 09:02
Bunuel wrote:

In the figure, if line CE bisects ∠ACB, then x =

(A) 45

(B) 50

(C) 55

(D) 65

(E) 70

Attachment:
triangle.jpg

40 + 90 + angle ACB = 180

Angle ACB = 50

since CE bisects angle ACB then we know that ACE = 25 and ECB = 25

Sum of angles in triangle ECB will be x + 25 + 90 = 180

x + 115 = 180

x = 65

Re: In the figure, if line CE bisects ∠ACB, then x =   [#permalink] 15 Oct 2018, 09:02
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