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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.

triangle

EDIT:enter image description here

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Isn't it easier to use the "Intersecting chords theorem"? enter image description here

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$\angle BAD = 90^{\circ} - \angle CBA = \angle ACD \implies \angle BAD \cong \angle ACD \implies \tan \angle BAD = \tan \angle ACD\implies \dfrac{BD}{AD} = \dfrac{AD}{DC}\implies AD^2 = BD\cdot DC$, which is what you are trying to prove.

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  • $\begingroup$ Sorry i meant Using the tools Euclid had at his disposal he didnt have distance let alone sin and cos :P i am fairly certain it can be done using two circles but im not sure how. $\endgroup$ – Faust Oct 14 '17 at 5:42
  • $\begingroup$ You would use Mike's solution below... as it fits the curriculum at your school..The sin and cos is small deal. Why reject it? $\endgroup$ – DeepSea Oct 14 '17 at 5:44
  • $\begingroup$ The book that i am working on is Euclidean geometry everything is done using Euclids propositions mostly done with a straight edge and compass. its rather fun to try and solve the problems in such way? (im not the one who downvoted but i can't begin to interpret what a cyclic ABEC is) $\endgroup$ – Faust Oct 14 '17 at 5:47
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I think that this is the kind of proof the OP is looking for

$Q_1+Q_2\doteq Q_3 \text{ Pythagoras}\\ Q_3\doteq Q_2+R \text{ Euclid lemma}\\ R+Q_2\doteq Q_1+Q_2 \text { transitive property of equivalence}\\ R\doteq Q_1$

enter image description here

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  • $\begingroup$ I think it' nothing. It's just tautology. $\endgroup$ – Michael Rozenberg Oct 14 '17 at 13:51
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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.

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  • $\begingroup$ Sorry what does "If we'll take $\Delta EBC\cong\Delta ABC$ such that $AE\cap BC=\{D\}$" mean exactly? Also what Object is ABEC and what does it mean to cyclic? $\endgroup$ – Faust Oct 14 '17 at 5:48
  • $\begingroup$ @Faust It means just take this! Also, $AE\cap BC=\{D\}$ means that $D$ is a common point of $AE$ and $BC$. Also, $ABEC$ is cyclic because $\measuredangle BAC+\measuredangle BEC=180^{\circ}.$ $\endgroup$ – Michael Rozenberg Oct 14 '17 at 5:59
  • $\begingroup$ Yeah i defiantly Havent learned anything about quadrilaterals yet in my textbook so there must be some more elementary means that i will have to figure out on my own. thanks for your help. $\endgroup$ – Faust Oct 14 '17 at 6:13
  • $\begingroup$ @Faust You are welcome! $\endgroup$ – Michael Rozenberg Oct 14 '17 at 6:15
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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.

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