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

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

and:
 * $KA^2 = KC^2 + CA^2$

Therefore:
 * $HA^2 = HK^2 + KC^2 + CA^2$

But from :
 * $HC^2 = HK^2 + KC^2$

Therefore:
 * $HA^2 = HC^2 + CA^2$

Therefore from :
 * $\angle HCA$ 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 :
 * $\triangle MDF = \triangle HAC$

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 :
 * $\triangle CAB = \triangle FDE$

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 :
 * $\triangle BCK = \triangle EFN$

Therefore:
 * $CK = FN$

But we have that:
 * $AC = DF$

Therefore from :
 * $\triangle ACK = \triangle DFN$

and so:
 * $AK = DN$

We have that:
 * $AH = DM$

and so:
 * $AH^2 = DM^2$

But $\angle AKH$ is a right angle.

So from :
 * $AK^2 + KH^2 = AH^2$

Also $\angle DNM$ is a right angle.

So from :
 * $DM^2 + DN^2 = NM^2$

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 :
 * $\angle HAK = \angle MDN$