prerequisites:
Isosceles triangle theorem (proof)
Proof:
Let \triangle ABC be a triangle with the given side lengths. If possible, let \triangle A'BC be another triangle such that A'B = AB and A'C = AC. Join the points A and A' by a straight line.
Now in \triangle ABA',
\qquad\quad AB = A'B
Hence \triangle ABA' is an isosceles triangle. Thus, by the isosceles triangle theorem,
\qquad\quad\!\!\angle BAA' = \angle BA'A
\therefore\quad\;\;\alpha + \beta = \delta
\therefore\quad\;\;\delta > \beta\qquad\qquad\qquad\qquad\qquad\cdots\text{(1)}
Similarly, in \triangle ACA',
\qquad\quad AC = A'C
Hence \triangle ACA' is an isosceles triangle. Thus, by the isosceles triangle theorem,
\qquad\quad\!\!\angle CAA' = \angle CA'A
\therefore\quad\;\;\beta = \delta + \gamma
\therefore\quad\;\;\delta < \beta
But this contradicts (1).
Hence, it is not possible to construct a distinct triangle \triangle A'BC.
Thus, the triangle \triangle ABC is unique.
Recommended:
Unique SAS triangle
Unique RHS triangle
SSS congruence
No comments:
Post a Comment