Condition for Equal Angles contained by Elevated Straight Lines from Plane Angles

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Theorem

In the words of Euclid:

If there be two equal plane angles, and on their vertices there be set up elevated straight lines containing equal angles with the original straight lines respectively, if on the elevated straight lines points be taken at random and perpendiculars be drawn from them to the planes in which the original angles are, and if from the points so arising in the planes straight lines be joined to the vertices of the original angles, they will contain, with the elevated straight lines, equal angles.

(The Elements: Book $\text{XI}$: Proposition $35$)


Porism

In the words of Euclid:

From this it is manifest that, if there be two equal plane angles, and if there be set up on them elevated straight lines which are equal and contain equal angles with the original straight lines respectively, the perpendiculars drawn from their extremities to the planes in which are the original angles are equal to one another.

(The Elements: Book $\text{XI}$: Proposition $35$ : Porism)


Proof

Euclid-XI-35.png

Let the angles $\angle BAC$ and $\angle EDF$ be equal.

From the points $A$ and $D$ let the elevated straight lines $AG$ and $DM$ be set up so they contain equal angles with the original straight lines:

$\angle MDE = \angle GAB$
$\angle MDF = \angle GAC$

Let $G$ and $M$ be arbitrary points on $AG$ and $DM$.

Let $GL$ and $MN$ be drawn from $G$ and $M$ perpendicular to the planes holding $\angle BAC$ and $\angle EDF$ respectively, meeting them at $L$ and $N$.

Let $LA$ and $ND$ be joined.

It is to be demonstrated that $\angle GAL = \angle MDN$.


Let $AH = DM$.

Let $HK$ be drawn through $H$ parallel to $GL$.

But $GL$ is perpendicular to the plane holding $\angle BAC$.

Therefore from Proposition $8$ of Book $\text{XI} $: Line Parallel to Perpendicular Line to Plane is Perpendicular to Same Plane:

$HK$ is perpendicular to the plane holding $\angle BAC$.


From the points $K, N$ let $KN, NF, KB, NE$ be drawn perpendicular to the straight lines $AC, DF, AB, DE$.

Let $HC, CB, MF, FE$ be joined.

From Proposition $47$ of Book $\text{I} $: Pythagoras's Theorem:

$HA^2 = HK^2 + KA^2$

and:

$KA^2 = KC^2 + CA^2$

Therefore:

$HA^2 = HK^2 + KC^2 + CA^2$

But from Proposition $47$ of Book $\text{I} $: Pythagoras's Theorem:

$HC^2 = HK^2 + KC^2$

Therefore:

$HA^2 = HC^2 + CA^2$

Therefore from Proposition $48$ of Book $\text{I} $: Square equals Sum of Squares implies Right Triangle:

$\angle HCA$ is a right angle.

For the same reason:

$\angle DFM$ is a right angle.

Therefore:

$\angle HCA = \angle DFM$

But:

$\angle HAC = \angle MDF$

Therefore from Proposition $26$ of Book $\text{I} $: Triangle Side-Angle-Angle Congruence:

$\triangle MDF = \triangle HAC$

Therefore:

$AC = DF$


Similarly it can be proved that $AB = DE$.


So we have that:

$AC = DF$
$AB = DE$

We also have that $\angle CAB = \angle FDE$

So from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

$\triangle CAB = \triangle FDE$

and so:

$\angle ACB = \angle DFE$

But the right angle $\angle ACK$ equals the right angle $\angle DFN$.

Therefore the remaining angles are equal:

$\angle BCK = \angle EFN$

For the same reason:

$\angle CBK = \angle FEN$

Therefore from Proposition $26$ of Book $\text{I} $: Triangle Side-Angle-Angle Congruence:

$\triangle BCK = \triangle EFN$

Therefore:

$CK = FN$

But we have that:

$AC = DF$

Therefore from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

$\triangle ACK = \triangle DFN$

and so:

$AK = DN$

We have that:

$AH = DM$

and so:

$AH^2 = DM^2$

But $\angle AKH$ is a right angle.

So from Proposition $47$ of Book $\text{I} $: Pythagoras's Theorem:

$AK^2 + KH^2 = AH^2$

Also $\angle DNM$ is a right angle.

So from Proposition $47$ of Book $\text{I} $: Pythagoras's Theorem:

$DM^2 + DN^2 = NM^2$

But as:

$AK = DN$

it follows that:

$AK^2 = DN^2$

Therefore:

$KH^2 = NM^2$

and so:

$KH = NM$

So we have that:

$AH = MD$
$AK = DN$
$HK = MN$

So from Proposition $8$ of Book $\text{I} $: Triangle Side-Side-Side Congruence:

$\angle HAK = \angle MDN$

$\blacksquare$


Historical Note

This proof is Proposition $35$ of Book $\text{XI}$ of Euclid's The Elements.


Sources

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