Comments on Similar triangles with the same area
Parent
Similar triangles with the same area
My kid was assigned this problem:
Given
$\overline{AC}\cong\overline{AB}$ and $\angle B\cong\angle C$.
Prove (a) $\overline{CE}\cong\overline{BD}$ and (b) $\overline{OB}\cong\overline{OC}$.
(I apologize for the ugly sketch. All lines that seem like they're meant to be straight are.)
The most direct way seems to be: Prove triangles $CAD$ and $BAE$ congruent (easy). Prove triangles $COE$ and $BOD$ similar (easy). Prove triangles $COE$ and $BOD$ congruent because they're similar and have the same area (by subtracting the common area of $ADOE$). The rest follows immediately.
The problem is — I don't recall ever seeing in a high-school geometry text the proposition "If two triangles are similar and have the same area, they're congruent". Or at least not as a method of proof for proving triangles congruent. So I guess my question is: Why not? It would certainly come in handy for cases like this. (Or am I wrong that it's correct in general?)
As a secondary question, is there a better way to do the problem above?
Post
For a triangle defined by three points $X, Y, Z$:
$$A_{\triangle XYZ}=\frac12\cdot\overline{XY}\cdot\overline{YZ}\cdot\sin{\angle Y}$$
Since $A_{\triangle ACD}=A_{\triangle AEB}$,
$$\overline{AB}\cdot\overline{BE}\cdot\sin\angle B=\overline{AC}\cdot\overline{CD}\cdot\sin\angle C$$
Since $\overline{AB}\cong\overline{AC}$ and $\angle B\cong\angle C$, this equation simplifies to
$$\overline{BE}=\overline{CD}$$
From there congruence of the triangles follows from SAS, and the rest follows as you set out in your original post. $\square$
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