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?
2 answers
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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. $\square$
One can easily prove the following fact:
If two triangles are similar, then the ratio of their areas is the square of the ratio of their sides.
Now, if two similar triangles have the same area, it follows that the ratio of the corresponding sides must be the square root of $1$, which is $1$. So, the triangles must be congruent.
Alternative Solution
Since $\overline{AB} \cong \overline{AC}$, $\angle B \cong \angle C$, and $\angle A$ is common, two triangles $\triangle ABE$ and $\triangle ACD$ are congruent. So we have $$\overline{AD} \cong \overline{AE} \quad \Rightarrow \quad |\overline{BD}|=|\overline{AB}|-|\overline{AD}|=|\overline{AC}|-|\overline{AE}|=|\overline{CE}| \quad \Rightarrow \quad \overline{BD} \cong \overline{CE}$$ (Please note that we used the assumption $\overline{AB} \cong \overline{AC}$ in the above).
Now, it can be easily followed that the two triangles $\triangle BOD$ and $\triangle COE$ are congruent. So, $$\overline{OB} \cong \overline{OC}.$$
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