Let $\triangle ABC$ have a right or obtuse angle at $\angle A$. Prove that any other triangle with the same $\angle A$ and longer sides $B,C$ will have a longer side opposite $A$.
My proof is below. I request verification, critique, or feedback. Note that this statement is less obvious than it might seem, and is in fact generally not true for acute triangles!
Let points $A, B, B'$ be colinear in that order (such that $AB < AB'$) and $A, C, C'$ be colinear in that order, and $\angle BAC$ be right or obtuse. We claim that $B'C' > BC$.
Observe that since $\angle ABC$ is acute, $\angle CBB'$, which is supplementary, must be obtuse. Consequently, $\angle CBB'$ is the greatest angle in $\triangle CBB'$, and $CB' > CB$.
Similarly, $\angle ACB'$ is acute, so $\angle C'CB'$ is obtuse, and therefore $C'B' > CB'$. Thus, $C'B' > CB$, QED.
Best Answer
Your proof looks correct, and is well presented. My only (rather minor) suggestion is that having a diagram would have made it a bit easier to follow.
FYI, here's an alternate proof method. In $\triangle ABC$, let the side lengths opposite $A$, $B$ and $C$ be $a$, $b$ and $c$, respectively. The law of cosines states that
$$a^2 = b^2 + c^2 - 2bc\cos(A) \tag{1}\label{eq1A}$$
Also, in $\triangle AB'C'$, have the side lengths opposite $A$, $B'$ and $C'$ be $a'$, $b'$ and $c'$, respectively. This then gives
$$(a')^2 = (b')^2 + (c')^2 - 2(b')(c')\cos(A) \tag{2}\label{eq2A}$$
Note $b' \gt b \;\;\to\;\; (b')^2 \gt b^2$ and $c' \gt c \;\;\to\;\; (c')^2 \gt c^2$. Also, since $\measuredangle A$ is a right or obtuse angle, then $\cos(A) \le 0 \;\;\to\;\; -\cos(A) \ge 0$, so $- 2(b')(c')\cos(A) \ge - 2bc\cos(A)$. Thus, the RHS of \eqref{eq2A} is greater than the RHS of \eqref{eq1A}, which means $(a')^2 \gt a^2 \;\;\to\;\; a' \gt a$.
Here's a second alternative proof method. With $\triangle ABC$, using the same side lengths as with \eqref{eq1A}, the law of sines says that
$$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \tag{3}\label{eq3A}$$
and, similarly with $\triangle AB'C'$, we have
$$\frac{a'}{\sin(A)} = \frac{b'}{\sin(B')} = \frac{c'}{\sin(C')} \tag{4}\label{eq4A}$$
Note the angles $B'$ and $C'$, compared to $B$ and $C$, respectively, have either not changed, or one of them has decreased and the other increased. WLOG, have $\measuredangle B' \le \measuredangle B$. Since $\sin(x)$ is a strictly increasing function for $0^{\circ} \lt x \lt 90^{\circ}$, then $\sin(B') \le \sin(B) \;\to\; \frac{1}{\sin(B')} \ge \frac{1}{\sin(B)}$. Since $b' \gt b'$, then comparing \eqref{eq4A} to \eqref{eq3A}, we get $\frac{b'}{\sin(B')} \gt \frac{b}{\sin(B)} \;\to\; \frac{a'}{\sin(A)} \gt \frac{a}{\sin(A)} \;\to\; a' \gt a$.