I'm studying Euclid's Elements to help me with my geometry skills, but I'm stuck with the language in B6.P8. English is not my first language, so that doesn't help, but maybe you guys can.
It says that if in a right-angled triangle, a line perpendicular to the base BC and through the right-angle A is drawn, then all the triangles are similar to the whole and one another. What I don't get in the proof is the difference between equiangular and similar. The first part is as follows:
Since $<BAC = <ADB$ (both 90 degrees) and $<B$ is common in $\triangle ABC, ABD$, it follows that $<ACB = <BAD$ (remaining angle). Thus, triangle ABC is EQUIANGULAR to triangle ABD.
Thus, as BC (subtending the right-angle in $\triangle ABC$) is to BA (subtending the right-angle in $\triangle ABD$),
so the same AB (subtending <C in $\triangle ABC$) is to BD (subtending <BAD in $\triangle ABD$),
and further AC is to BD (both subtending <B)
Thus, $\triangle ABC$ is equiangular to $\triangle ABD$ and has the sides about the equal angles proportional. Thus, $\triangle ABC$ is SIMILAR to $\triangle ABD$ (and similarly we can show that $\triangle ABC$ is similar to $\triangle ADC$.
It is then left to show that $\triangle ABD$ is similar to $\triangle ADC$, but that's done in an equivalent way. What I struggle with is the difference between equiangular and similar. I believed that equiangular meant that all angles are equal (thus all 60 degrees in a triangle) but that does not uphold here, as we have an angle of 90 degrees and two "arbitrary" other ones...
Please help :') Thanks!
In Euclidean geometry, when we say two triangles are "equiangular," it means that they have the same angles (alternatively, their corresponding angles match, following the comment by @lulu). In other words, all the corresponding angles in both triangles are equal. Equiangular triangles do not necessarily have to have angles of $60$ degrees each; they can have any set of angles as long as they match in both triangles being compared. In your case, the key angles are the right angles at points $A$ in triangles $ABC$ and $ABD$. These right angles are equal (both $90$ degrees), which is why we say that triangle $ABC$ is equiangular to triangle $ABD$.
Now, when we say two triangles are "similar," it means that their corresponding sides are in proportion, and their corresponding angles are equal. So, while equiangular triangles have equal angles, similar triangles not only have equal angles but also their side lengths are proportional.
In your proof, you've already established that triangle $ABC$ is equiangular to triangle $ABD$ because they share the same right angle and one additional angle. This establishes that they have equal angles.
Then, by the reasoning provided in your proof, you've shown that the sides of triangle $ABC$ are in proportion to the sides of triangle $ABD$. In particular, you mentioned that $BC$ is to $BA$ as $AB$ is to $BD$, and so on. This proportionality of sides confirms that triangle $ABC$ is similar to triangle $ABD$.
So, in summary: