vittovangind wrote:
Is triangle ABC an isosceles triangle?
(1) AD = DC
(2) The area of triangle ABD equals the area of triangle BDC
NOTE: I added labels v, w, x, y, and z to make it easier to reference the sides.
Target question: Is triangle ABC an isosceles triangle?
To show that triangle ABC is an isosceles triangle, we can either show that the triangle has two equal angles or two equal sides.
Statement 1: AD = DC (i.e., x = y)
There's a nice rule that says that the altitude of an isosceles triangle always bisects the opposite side. Since the altitude (z) bisects AC, we can conclude that
triangle ABC is an isosceles triangle.
Since we can answer the
target question with certainty, statement 1 is SUFFICIENT
Here's
another way to show that statement 1 is sufficient.
Notice that the two smaller triangles that comprise triangle ABC are both RIGHT TRIANGLES.
The two triangles both SHARE SIDE z, and statement 1 tells us that side x = side y.
Since both smaller triangles are right triangles, we COULD use the Pythagorean Theorem to determine the hypotenuses (v and w) of the two triangles. MOREOVER, since we'd be plugging the SAME NUMBERS into the Pythagorean Theorem, the hypotenuses (v and w) will be the same length.
If v and w have the same length, then
triangle ABC is an isosceles triangle.
Since we can answer the
target question with certainty, statement 1 is SUFFICIENT
Statement 2: area of triangle ABD = area of triangle BDC
Area of triangle = (1/2)(base)(height)
So, (1/2)(x)(z) = (1/2)(y)(z)
If we divide both sides by (1/2)(z), we see that x = y
Since statement 1 also told us that x = y, and since we already determined that statement 1 is sufficient, we can conclude that statement 2 is SUFFICIENT
Answer =
D
Cheers,
Brent
