Stupidly simple geometry problem I can't do

Question

Okay. Here it goes.

C and D are two points on the same side of a straight line AB and P is any point on AB. Show that PC + PD is least when the angles CPA and DPB are equal.

I have no idea why I can't do this. I've drawn diagrams, tried using the triangle inequality and it still escapes me.

A hint would be nice just so I can put this question out of its misery.

score 7 · Accepted Answer · answered Oct 12 '14 at 03:40

7

Your problem is "What is the shortest path from $C$ to $D$ that touches line $AB$?". Reflect $D$ in the line $AB$ and obtain $D'$. Now, $CP+PD=CP+PD'$, and your problem becomes "What is the shortest path from $C$ to $D'$ that touches line $AB$?", which looks quite similar to the previous one except for the fact that now every path from $C$ to $D'$ touches line $AB$ ...

answered Oct 12 '14 at 03:40

wisefool

4,193
13
14

Oh god. All along, I had been reading the question as "show that PC + PD is least when CDP is isosceles." – Soham Chowdhury Oct 12 '14 at 03:54

user175755 · Answer 2 · 2014-10-12T04:19:12.120

3

C' is symmetric to C with regard to AB, hence triangles FCP=FC'P and so angle CPF=FPC' Because angle FPC'=DPB we get that angle CPA=DPB. Distance DP+PC'=DC'=CP+PD, since the shortest distance between two points is a line we get that CP+PD is shortest.

edited Oct 12 '14 at 04:19

answered Oct 12 '14 at 04:08

user175755

163

2

The picture would be even better if C and D were not about the same distance from the line AB – Henry Oct 12 '14 at 06:50
It doesn't change anything. Method of construction works for any arbitrary points. :) – user175755 Oct 12 '14 at 12:30

Stupidly simple geometry problem I can't do

2 Answers2