If $P$ is any point on a straight line drawn through the vertex $A$ of an isosceles triangle $ABC$, parallel to the base, prove that $PB+PC>AB+AC$

Question

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

I am not able to figure out as to how we can prove this. It would be great if someone could help.

...but you've answered your own question! – Parcly Taxel Sep 23 '16 at 16:11 — Parcly Taxel, Sep 23 '16 at 16:11
but i dont know how to prove the latter – Ishita Hariramani Sep 23 '16 at 16:14 — Ishita Hariramani, Sep 23 '16 at 16:14

score 7 · Accepted Answer · answered Sep 23 '16 at 16:26

7

Reflect $C$ at the line $AP$ to obtain $C'$. Then $AB+AC=BA+AC'$, where $B,A,C'$ are colinear, whereas $PA+PC=AP+PC'$, where $BPC'$ are not collinear. Hence this is just an example of: The straight line from $A$ to $C'$ is the shortest.

answered Sep 23 '16 at 16:26

Hagen von Eitzen

374,180

great! just that it would be PB+PC=BP +PC' in 2nd line – Ishita Hariramani Sep 24 '16 at 07:17

If $P$ is any point on a straight line drawn through the vertex $A$ of an isosceles triangle $ABC$, parallel to the base, prove that $PB+PC>AB+AC$

1 Answers1