In Delta XYZ,XY XZ and P is any point on th | vedclass

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(N/A) In $\Delta XYZ$,$XY > XZ$ (Given).$	herefore \angle XZY > \angle XYZ$ (Since the angle opposite to the larger side of a triangle is greater).In

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(N/A) In $\Delta XYZ$,$XY > XZ$ (Given).
$\therefore \angle XZY > \angle XYZ$ (Since the angle opposite to the larger side of a triangle is greater).
In $\Delta XPZ$,$\angle XPY$ is an exterior angle,and $\angle XZP$ is its interior opposite angle.
Therefore,$\angle XPY > \angle XZP$ (Exterior angle property).
Since $\angle XZY$ is the same as $\angle XZP$,we have $\angle XPY > \angle XZP > \angle XYZ$.
Thus,$\angle XPY > \angle XYZ$ (or $\angle XPY > \angle XYP$).
In $\Delta XYP$,since the angle opposite to $XY$ (which is $\angle XPY$) is greater than the angle opposite to $XP$ (which is $\angle XYP$),it follows that $XY > XP$.

In $\Delta XYZ$,$XY > XZ$ and $P$ is any point on the side $YZ$. Prove that $XY > XP$.

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