What is the measure of angle DCB?

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In the following figure, $AD = AB$. Also $\angle DAB = \angle DCB = \angle AEC = 90^\circ$ and $AE = 5$. Find the area of quadrilateral $ABCD$.

What I Tried: Here is the figure :-

I could conclude that $ABCD$ is cyclic, but I really could not use that property in a useful way till now. I thought about Ptolemy's Theorem but I am not sure since I don't know enough lengths.

Instead, Pythagoras Theorem gives the required values, assuming $AD = AB = x$ . But I am still stuck as that is not enough for finding the side-lengths of the quadrilateral, only then I can find it's area. The red angles came to be $45^\circ$, and I am stuck right here. I can't seem to use something with the cyclic quadrilaterals.

Can anyone help me? Thank You.

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In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60 [#permalink]

01 Jul 2015, 02:20

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In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60° and angle ACD is 20°, what is the measure of angle x?A. 10°B. 15°C. 20°D. 30°E. 40°

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Re: In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60 [#permalink]

01 Jul 2015, 02:27

Bunuel wrote:

In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60° and angle ACD is 20°, what is the measure of angle x?A. 10°B. 15°C. 20°D. 30°E. 40°

Kudos for a correct solution.

The Answer Option A

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Re: In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60 [#permalink]

01 Jul 2015, 02:22

Bunuel wrote:

In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60° and angle ACD is 20°, what is the measure of angle x?A. 10°B. 15°C. 20°D. 30°E. 40°

Kudos for a correct solution.

since all three are isosceles triangle 2(x+20+60)=180.. x=10 ans A _________________

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Re: In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60 [#permalink]

06 Jul 2015, 01:33

Bunuel wrote:

In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60° and angle ACD is 20°, what is the measure of angle x?A. 10°B. 15°C. 20°D. 30°E. 40°

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

If AD = DB = DC, then the three triangular regions in this figure are all isosceles triangles. Therefore, you can fill in some of the missing angle measurements as shown to the right. Since you know that there are 180° in the large triangle ABC, you can write the following equation:x + x + 20 + 20 + 60 + 60 = 1802x + 160 = 180x = 10Answer: A. _________________

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Re: In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60 [#permalink]

10 Jul 2015, 06:16

Hi there, I tried to make a solution easier, but don't get why do I get the wrong answer. Could someone shade some light on it...Geometry is not my best topic in math....

If you just connect a line from A to BC (let say Point Y) so we get a right angle 90. Then we have to sum up all the angles (Trianlge AYB): 90+60+x+x=180 , X=15

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In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60 [#permalink]

10 Jul 2015, 06:23

BrainLab wrote:

Hi there, I tried to make a solution easier, but don't get why do I get the wrong answer. Could someone shade some light on it...Geometry is not my best topic in math....

If you just connect a line from A to BC (let say Point Y) so we get a right angle 90. Then we have to sum up all the angles (Trianlge AYB): 90+60+x+x=180 , X=15

No, you are incorrect. Let me show you why. If we do create a right angled triangle AYB with 90 degree at Y and B = (60+X) degree (because DB = DC in triangle DBC and AD = BD in triangle ADB), you will not get Angle (YAB) =x, as for this to happen , you need to have triangle ABC as iscosceles (or equilateral). This is because, you can not just extend AD to meet B at 90 degree to BC if the triangle is not Iscosceles (or equilateral). NOwehere in the question is it mentioned that triangle ABC is iscosceles etc.In the given question, if you look at the solutions above, you will see Angle (ABC) = 70 while Angle (ACB) = 80. Thus triangle ABC is not iscosceles. This is where you are making a mistake. You are assuming that triangle ABC is iscosceles (without this assumption, you can not extend AD to meet BC at 90 degrees.). Hope this clears your doubt.

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Re: In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60 [#permalink]

10 Jul 2015, 23:01

Hi Engr2012, thanks... In other words I got trapped by the picture ))... and we can not assume that Line BC is straight without any slope.

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In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60 [#permalink]

11 Jul 2015, 04:02

BrainLab wrote:

Hi Engr2012, thanks... In other words I got trapped by the picture ))... and we can not assume that Line BC is straight without any slope.

Sorry didn't get what you wanted to say "we can not assume that Line BC is straight without any slope." BC will be a line no matter what the slope is. The issue at hand is whether AD when extended to BC, intersects BC at right angles. AD will be intersecting BC at right angles ONLY IF the triangle is isosceles or equilateral. _________________

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Re: In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60 [#permalink]

06 Apr 2021, 21:26

Since it is given that in triangle ABC , AD=DB=DC , it implies that D is the circumcenter of the triangle and all the points A,B,C lie on a circle passing through them.now in triangle DCB , DC=DB and angle DCB=60. so angle DBC= angle CDB= 60. now angle BAC is an inscribed angle, so it is half of the central angle.so angle BAC = 30. angle CAD = angle DCA= 20.

SO, x=30-20=10.

Re: In Triangle ABC, AD = DB = DC (see figure). Given that angle DCB is 60 [#permalink]

06 Apr 2021, 21:26