How can you show that if two sides of a triangle are not congruent, then the angles opposite those sides are not congruent, and the larger angle is opposite the larger side of the triangle. [clue: Use isosceles triangles.]

  • $\begingroup$ Is it a homework problem? $\endgroup$ – Yimin Mar 11 '13 at 15:13

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Clearly, $\angle ABC>\angle ACB$

Let $D$ be a point on $AC$ such that $\angle ABD=\angle ADB$

Also let $E$ be the midpoint of $BD$

So, in $\triangle ABE, \triangle ADE$

$(1) BE=DE,(2) AE$ is the common side and $(3) \angle ABE=\angle ADE$

So, $\triangle ABE, \triangle ADE$ are congruent (Using $SAS$ formula)

So, $AB=AD$ and $\angle AEB=\angle AED=\frac{\angle AEB+\angle AED}2=\frac\pi2$

Using Pythagoras Theorem, $AC^2=AE^2+EC^2,AD^2=AE^2+ED^2\implies AC^2-AD^2=EC^2-ED^2>0$

So, $AC>AD\implies AC>AB$

  • $\begingroup$ Great job! I draw a triangle to check your answer and it answered all what is required. =) $\endgroup$ – caleb Mar 11 '13 at 16:37
  • $\begingroup$ @caleb, hope I could make the idea clear. $\endgroup$ – lab bhattacharjee Mar 11 '13 at 16:42

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