Princ wrote:
Attachment:
Capture.PNG
In the figure above, if \(\triangle\) ABC and \(\triangle\) ADC are isosceles triangles, what is the value of x\(^{\circ}\)?
(1) y = 140\(^{\circ}\)
(2) \(\angle\) DCA is one half the measure of \(\angle\) BCA.
Neat question!
Jot down that the triangles are isosceles.
The problem doesn't have the 'figures not drawn to scale' text. However, in DS, you should always assume that figures may not be drawn to scale. This is a common trap with isosceles triangles. They've drawn them so that it looks like AB and BC are equal, and it looks like AD and DC are equal.
But you don't know whether that's actually true! It could be the case that AD and AC are the equal sides, or BC and AC. Be very cautious working with the statements.
(1) y = 140 degrees. This is insufficient. Imagine 'stretching' the triangle - you can change the value of x substantially, while keeping y equal to 140. Since x can change, this statement isn't sufficient.
(2) DCA is one half of BCA. This looks insufficient, but let's just make sure. Here are two different setups with isosceles triangles, where DCA is one half of BCA, but where x has different values.
(1 and 2) Our goal is to prove insufficiency. We want to show that there are two different ways to do the drawing, where both statements are true, but where x has two different values. Here's the easiest one:
Let's see if we can play with the 'isosceles triangle' thing and try to come up with a different triangle. Suppose in the smaller triangle, y = 140, and both smaller angles are 20. But what if the outer triangle looked different?
This drawing still follows all of the rules! Both triangles are isosceles, y = 140, and DCA is half of BCA. But x has a different value.
So,
1+2 is insufficient. The answer is
E.