wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let P be any point on circumference of triangle ABC and perpendicualr PL, PM, PN are drawn on the segments BC, CA and AB respectively. Show points L,M, N are collinear.

Open in App
Solution

To prove : N, M and L are collinear.
Construction: Join PA and PC.
Proof: PMA+ANP=90+90=180
Therefore, points A, M, P, N are conyclic.
PMN=PAN ....(i)
(angles in the same segment)
Again PMC=PLC=90
Therefore quadrilateral PMLC is cyclic.
PML+PCL=180 .....(ii)
Now, PAB+PCB=180
and PAN+PAB=180
PAN=PCB=PCL ....(iii)
From Equations (i) (ii) and (iii) we get,
PML+PCL=PML+PAN
=PML+PMN=180
Hence, N, M, L are collinear. (NML line is called simpson's line).

flag
Suggest Corrections
thumbs-up
15
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Vertically Opposite Angles
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon