Let ABC be a right triangle, and let AD be the altitude from the right angle A to the hypotenuse BC. Prove that $AD^2 = BD $ x $ DC$ (in the sense of content). I believe what this is asking us to show is that the square on $AD $ x $AD $ has the same content as the rectangle on $BD $ x $ DC $ i do know how to square a given rectangle but only now how to show its the same as a given square.
EDIT: Any ideas how to do this proof using the tools Euclid had at his disposal.
EDIT:
On
Use $$\Delta ABD\sim\Delta CAD.$$
Why $AD^2$ it's an area of the square only.
If we'll take $\Delta EBC\cong\Delta ABC$ such that $AE\cap BC=\{D\}$,
then $ABEC$ is cyclic and we obtain: $$AD\cdot DE=BD\cdot DC.$$
We got $$AD\cdot DE=AD^2,$$ which is the square again.
On
$\angle BAD = 90^{\circ} - \angle CBA = \angle ACD \implies \angle BAD \cong \angle ACD \implies \tan \angle BAD = \tan \angle ACD\implies \dfrac{BD}{AD} = \dfrac{AD}{DC}\implies AD^2 = BD\cdot DC$, which is what you are trying to prove.
On
I think that this is the kind of proof the OP is looking for
$Q_1+Q_2\doteq Q_3 \text{ Pythagoras}\\ Q_3\doteq Q_2+R \text{ Euclid lemma}\\ R+Q_2\doteq Q_1+Q_2 \text { transitive property of equivalence}\\ R\doteq Q_1$
On
Though @Raffaele is Correct i just wanted to elaborate on proving this directly there are 2 ways i have found.
First note by III.31 (Euclid book 3 prop 31) Thaeles theorem states that if we circumscribe the right traingle not only does A,B,C lie on the circle but the side BC is the diameter of the Circle.
1)From here you may extend the rectangle down from AD and show that the method of squaring a circle yields the square on AD.
2) you may find the midpoint of BC call it O then jion OA this is a radius since A lies on the circle, noting that you then make an argument on the inner triangle using BD= BC-DC and noting that the midpoint (also the radius) is equal to one half of BD+DC Pythagoras theorem takes care of the rest.
Isn't it easier to use the "Intersecting chords theorem"?