Elementary Geometry

Miscellaneous Propositions

976

Theorem

When three perpendiculars to the sides of a triangle 𝐴𝐵𝐶, intersecting them in the points 𝑎, 𝑏, 𝑐 respectively, are concurrent, the following relation is satisfied; and conversely, if the relation be satisfied, the perpendiculars are concurrent. 𝐴𝑏²−𝑏𝐶²+𝐶𝑎²−𝑎𝐵²+𝐵𝑐²−𝑐𝐴²=0

Proof

If the perpendiculars meet in 𝑂, then 𝐴𝑏²−𝑏𝐶²=𝐴𝑂²−𝑂𝐶², ⋯ (I.47).

Examples

By the application of this theorem, the concurrence of the three perpendiculars is readily established in the following cases:

• When the perpendiculars bisect the sides of the triangle.
• When they pass through the vertices. (By employing I.47)
• The three radii of the eseribed circles of a triangel at the points of contact between the vertices are concurrent. So also are the radius of the inscribed circle at the point of contact with one side, and the radii of the two escribed circles of the remaining sides at the points of contact beyond the included angle.
  In these cases employ the values of the segments given in (953).
• The perpendiculars equidiculars from the vertices with three concurrent perpendiculars are also concurrent.
• When the three perpendiculars from the vertices of one triangle upon the sides of the other are concurrent, then the perpendiculars from the vertices of the second triangle upon the sides of the first are also concurrent.

Proof

If 𝐴, 𝐵, 𝐶 and 𝐴′, 𝐵′, 𝐶′ are corresponding vertices of the triangle, join 𝐴𝐵′, 𝐴𝐶′, 𝐵𝐶′, 𝐵𝐴′, 𝐶𝐴′, 𝐶𝐵′, and apply the theorem in conjunction with (I.47)

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive