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Let $ABC$ be a triangle in the plane and $X,Y$ and $Z$ points on the segments $BC,CA$ and $AB$. $X,Y$ and $Z$ are not identical to any corner of $ABC$.

Additionally, for a given $X$, let $P$ be the intersection of segment $CA$ and the line through $X$ parallel to $AB$. Finally, $Q$ shall be the intersection of $AB$ and the line through $X$ which is parallel to $CA$.

I want to prove the following two statements: For a given $X$ on $BC$,

  1. if point $Y$ is located between $P$ and $A$ and point $Z$ is located between $Q$ and $B$, then $A_{XYZ}>A_{XPZ}$. See picture
  2. if point $Y$ is located between $C$ and $P$ and point $Z$ is located between $A$ and $Q$, then $A_{XYZ}>A_{XPZ}$. See picture

The statements seem kind of clear to me geometrically if I compare the perpendicular heights of the two triangles ($Y$ has a bigger distance to line $ZX$ than $P$). But have you got any ideas about how to prove it?

Thank you in advance for your help!

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In case 1 (see picture) draw from $Y$ a line $YH$ parallel to $AB$. It follows from the hypotheses that $YH>PX$. But the heights of triangles $YXZ$ and $PXZ$ with respect to commmon base $AB$ (dashed lines in the picture) are proportional to $PX$ and $YH$, so the height of $YXZ$ is greater than the height of $PXZ$. It follows that the area of $YXZ$ is greater than the area of $PXZ$.

A similar argument holds for case 2.

enter image description here

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