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

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 = 180
2x + 160 = 180
x = 10

Answer: A.

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.

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.

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.