Post History
#2: Post edited
by
MathPhysics
·
2020-10-02T11:25:38Z (about 4 years ago)
- 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.
- 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: Initial revision
by
MathPhysics
·
2020-10-02T10:32:51Z (about 4 years ago)
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.