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A friend of mine posted this GRE geometry question and I gave it a look: https://i.stack.imgur.com/OwuhB.png (below) - I saw another question posted here which is similar to this but doesn't quite answer what I am asking.

enter image description here

The question is: is quantity A greater than quantity B, quantity B greater than quantity A, are they equal or is there not enough information to deduce this. And the only piece of information is the what's given below the figure: XY = YZ. The answer given is that they are equal.

From what I know if the areas are to be equal then their altitudes must be the same. WY can only be the altitude if XW = WZ (isosceles triangle). But this isn't stated. Can we deduce this from similarity principles? If so, how? If not, can we assume that the two triangles will have the same height because they share a common vertex?

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    $\begingroup$ $WY$ does not need to be an altitude (i.e. perpendicular to $XZ$) but if $XY=YZ$ then $WY$ is a median. Since the triangles $XYW$ and $YZW$ have the same perpendicular height and have equal bases, they must have the same area $\endgroup$
    – Henry
    Aug 4, 2017 at 23:01

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If it is given that $XY=YZ$, then the two triangles have the same height and the same base, so they must have equal areas, because the area of a triangle is half the product of its base and height.

The common height is simply the length of the perpendicular from $W$ to $XZ$, which may or may not coincide with $WY$. Note that in such questions, the figures are not necessarily drawn to scale, so you can only use the information provided by the words.

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The shared side $WY$ is not necessarily the altitude of either triangle, because we don't know if it is perpendicular to the base $XZ$.

However, the triangles do have a shared altitude -- it just might not be shown in the diagram. Consider the sketch below: enter image description here

In this diagram $XY$=$YZ$ and the dotted line is the altitude of both triangles. The altitude lies inside one triangle and outside the other, but is perpendicular to both of the two equal bases.

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  • $\begingroup$ A lot of good answers here! I'm selecting this one as the answer because of the diagram. I guess the trap I fell into was assuming WY was the altitude, and then trying to prove that assumption. $\endgroup$
    – dnclem
    Aug 4, 2017 at 23:13
  • $\begingroup$ BTW what program did you draw this diagram in? $\endgroup$
    – dnclem
    Aug 5, 2017 at 5:30
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    $\begingroup$ @david Geogebra. $\endgroup$
    – mweiss
    Aug 6, 2017 at 1:58
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Let $Y$ be the midpoint of $XZ$. Consider a line $XE$ perpendicular to $WY$, and a line $ZF$ perpendicular to $WY$. Since these lines are both perpendicular to $WY$, they are parallel to each other. Hence $\angle EXY \cong \angle FZY$. Moreover, $\angle XYE \cong \angle ZYF$, and $XY \cong YZ$ by hypothesis. Hence by the angle-side-angle congruence relation, we have $\triangle YXE \cong \triangle YZF$. In particular, $XE \cong ZF$, where these segments are the altitudes of the triangles $\Delta WYX$ and $\Delta WYZ$, respectively. As the triangles share a common base $WY$, and they have the same altitude, their areas are equal.

enter image description here

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  • $\begingroup$ Actually sorry that was an assumption I made. I am not sure if WY is an altitude (it's not stated). $\endgroup$
    – dnclem
    Aug 4, 2017 at 23:04
  • $\begingroup$ It might be helpful to edit the title of your question to reflect that... $\endgroup$
    – Xander Henderson
    Aug 4, 2017 at 23:08
  • $\begingroup$ To be fair I said "WY can only be the altitude if XW = WZ (isosceles triangle)", I didn't state that WY was indeed the altitude. But sorry for the ambiguity regardless. $\endgroup$
    – dnclem
    Aug 4, 2017 at 23:14

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