# 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

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, $$\overline{OB} \cong \overline{OC}.$$

#### 1 comment

Your alternative proof for the problem my kid had is a good one, and is probably the one intended by the problemsetter. Thank you!

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. $\square$

#### 1 comment

This I feel is a solid proof, but is probably not the one intended. I gather your kid is taking middle/high school Geometry and has not yet learned Trig, and certainly not this formulation of the area of a triangle.

## 1 comment

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