Triangles ABC, ACD, and ABD are all isosceles triangles. Point E (not

Question

https://gmatclub.com/forum/download/file.php?id=66005 Triangles ABC, ACD, and ABD are all isosceles triangles. Point E (not shown) is the midpoint of \overline{BD} . If the ratio of the length of \overline{CE} to the length of \overline{BC} is equal to \sqrt{3} : 2 , then what is the measure, in degrees, of ∠CAD? A. 10 B. 15 C. 30 D. 45 E. 60 1.jpg Are You Up For the Challenge: 700 Level Questions

Jasmeet07 · Answer

Can somebody explain this?

Dev23 · Answer

Triangle BCE is congruent to Triangle DCE ( BC=CD,BE=DE (E is midpoint of side BD), CE is common )
Therefore in angle BEC=angle DEC= 90 degree
In Triangle BCE side BE =1, BC =2 & CE = √3 (Pythagoras theorem)
Angle CBD=Angle CDB =60 
Angle BEC=angle DEC= 90 
Angle BCE = Angle DCE =30
Therefore Angle ACB= Angle ACD = 150 degree (Angle ACB+ACD+BCD=360)
Therefore in Triangle ACD
Angle CAD=Angle CDA = 15 ( Isosceles triangle)

arvind910619 · Answer

Line CE/BC =sqrt (3)/2

In triangle BCE

Cos O = Base/Hypotenuse

Cos O = sqrt (3)/2

Now according to trigonometry Cos 30 = sqrt (3)/2

So angle BCE = 30 deg

Now exterior angle property of the triangle 
Angle BCE = Angle BAC + Angle ABC 
Since Triangle ACB is a isosceles triangle

So Angle BAC = Angle ABC 
Therefore Angle BAC = Angle ABC = 15 deg

Same case is followed for triangle CAD and angle ∠CAD = 15 deg

Fdambro294 · Answer

Very tricky question. Although you can assume your way to what is likely the correct answer, which in this case it does turn out to be the right answer…

1st). The extended line from point C to point E on side BD bisects the side BD.

Thus, line AE is a Median of the isosceles triangle. However, we can not be sure which sides are the two equal sides and which side is the non equal base side.

(2nd) we are given that side CE : BC is in the ratio of: sqrt(3) : 2

Using the Trigonometric Ratios, the cosine of an angle (X) is determined by = (Opposite Side) / (Hypotenuse) of the right angled triangle containing the corresponding angles

In the case of the ratio: sqrt(3) : 2

The Cosine of (30 degrees) is equal to the ratio of the (opposite side) / (hypotenuse) in a 30-60-90 Right Triangle.

For this reason, angle <ECB must be = 30 degrees

From this point, you can determine the other angles using similar trigonometric ratios. The Angle at <E is 90 degrees and angle <EBC = 60 degrees

You can perform the same analysis on the other triangle created by line CE: triangle ECD is also a 30-60-90 degree right triangle

(3rd) at this point, we know that LINE AE drawn from vertex A is both:

The MEDIAN of side DB
And
The Altitude drawn to side DB

Therefore, line AE must be the line of symmetry for the Isosceles triangle, making the sides AD and AB the two equal sides of the Isosceles Triangle

However, we still do not now which sides are the two equal sides within the isosceles triangles ACB and ACD

At this point, you can observe that the exterior angle at point C creates a straight line angle with Line CE.

Since the Angle at <ECB = 30 degrees

The angles at: <DCA and <BCA must be 150 degrees

Since the base angles opposite the two equal sides must have the same measure, the 4 remaining unknown angles must be opposite the equal sides in each isosceles triangle. (Since we can not have two equal angles of 150 degrees in the same triangle

At this point, you can pick one of the isosceles triangles and solve for the interior angles.

Within triangle DCA:

Angle <DCA = 150 degrees 
Angle <DAC = X
Angle <CDA = X

X + X + 150 = 180

X = 15 degrees = angel <DCA

Answer:

15 degrees

Posted from my mobile device

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Triangles ABC, ACD, and ABD are all isosceles triangles. Point E (not

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Triangles ABC, ACD, and ABD are all isosceles triangles. Point E (not

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