Three perpendicular theorem

Content

Formulation of the theorem

If a straight line drawn on a plane through the base of an inclined one is perpendicular to its projection, then it is also perpendicular to the most inclined one.

Proof

Let be $A B$ ${\ displaystyle AB}$ - perpendicular to the plane $\alpha$ ${\ displaystyle \ alpha}$ , $A C$ ${\ displaystyle AC}$ - inclined and $c$ ${\ displaystyle c}$ - straight in the plane $\alpha$ ${\ displaystyle \ alpha}$ passing through the point $C$ ${\ displaystyle C}$ and perpendicular projection $B C$ ${\ displaystyle BC}$ . Draw a straight line $C K$ ${\ displaystyle CK}$ parallel straight $A B$ ${\ displaystyle AB}$ . Straight $C K$ ${\ displaystyle CK}$ perpendicular to the plane $\alpha$ ${\ displaystyle \ alpha}$ (as it is parallel $A B$ ${\ displaystyle AB}$ ), and therefore, any straight line of this plane, therefore $C K$ ${\ displaystyle CK}$ perpendicular to straight $c$ ${\ displaystyle c}$ . Draw through parallel lines $A B$ ${\ displaystyle AB}$ and $C K$ ${\ displaystyle CK}$ plane $\beta$ ${\ displaystyle \ beta}$ (parallel straight lines define a plane, and only one). Straight $c$ ${\ displaystyle c}$ perpendicular to two intersecting straight lines lying in the plane $\beta$ ${\ displaystyle \ beta}$ , this $B C$ ${\ displaystyle BC}$ by condition and $C K$ ${\ displaystyle CK}$ By construction, it means that it is perpendicular to any line belonging to this plane, which means that it is perpendicular to $A C$ ${\ displaystyle AC}$ .

Theorem, inverse to the three perpendicular theorem

If the straight line drawn on the plane through the base of the inclined is perpendicular to the most inclined, then it is perpendicular to its projection.

Proof

Suppose AB is perpendicular to the plane α , AC is inclined, and c is a straight line in the plane α passing through the base of inclined C. Let's draw a straight line SC , parallel to a straight line AB . The straight line SC is perpendicular to the plane α (according to this theorem, since it is parallel to AB ), and therefore any straight line of this plane, therefore, the SC is perpendicular to the line c . Let us draw the plane β through parallel straight lines AB and CK (parallel straight lines define a plane, and only one). A straight line with perpendicular to two straight lines lying in the plane β , this is the AU by condition and the CK according to the three perpendicular theorem, so it is perpendicular and any straight line belonging to this plane means a perpendicular to the straight line BC . In other words, the projection of BC is perpendicular to the straight line c lying in the plane α .

Usage Example

Prove that through any point of a straight line in space you can draw a straight line perpendicular to it.

Solution

Solution: let a be a straight line and A be a point on it. Take any point X outside the line a and draw through this point and the line a and the plane α . In the plane α through the point A it is possible to draw a straight line b , perpendicular to a .

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Reference tasks

Three perpendicular theorem

Content

Formulation of the theorem

Proof

Theorem, inverse to the three perpendicular theorem

Proof

Usage Example

Solution

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