Problem
GEO-B2-M05-P007 Image of a Line
#7
★★★☆☆ Level 3 of 5
Line \(l\) does not pass through \(O\). Let \(H\) be the foot of the perpendicular from \(O\) to \(l\), and let \(H'\) be the image of \(H\). Prove that the image of \(l\) lies on the circle with diameter \(OH'\).
For \(P\in l\), compare triangles \(OPH\) and \(OH'P'\).
Let \(P'\) be the image of \(P\). Since \(OP\cdot OP'=OH\cdot OH'=R^2\), we have \(\frac{OP}{OH'}=\frac{OH}{OP'}\). The angle at \(O\) is common, so \(\triangle OPH\sim\triangle OH'P'\). From \(\angle OHP=90^\circ\), it follows that \(\angle OP'H'=90^\circ\). Thus \(P'\) lies on the circle with diameter \(OH'\).
The key lemma of the module.