# A point inside a triangle

I am given a triangle $$\triangle ABC$$ with side lengths $$a,b,c$$ and a point $$P$$ inside it.
$$R_A=PA$$, $$R_C=PC$$, $$R_C=PC$$
the distances from point $$P$$ to the sides $$BC, AC, AB$$ are $$d_a, d_b, d_c$$ respectively.

How can I prove $$b\cdot d_a+a\cdot d_b \leq c\cdot R_C$$? I'd like to get a hint for where to start.

My attempt (I didn't find the solution but thats the closest I could get):

Firstly, I added points $$D,E,F$$ the projections of $$P$$ on sides $$BC,AC,AB$$ respectively. Also $$BC=a,AC=b,AB=c$$. Then: $$b\ge EC \space\space\space\space and \space\space\space\space a\ge DC$$ $$b\cdot d_a\ge EC\cdot d_a \space\space\space\space and \space\space\space\space a\cdot d_b \ge DC \cdot d_b$$ $$b\cdot d_a+a\cdot d_b \ge EC\cdot d_a+DC \cdot d_b=DE \cdot R_C$$

• I solved your problem. If you want to see my solution, show please your attempts. – Michael Rozenberg Jan 10 at 21:15
• @MichaelRozenberg I edited what I found into the original post, although it's not what I need to prove. Is it a good direction? – aradarbel10 Jan 10 at 22:06

## 2 Answers

The hint.

Make the following.

1. Prove that: $$ad_a+bd_b\leq cR_c;$$

2. Prove the previous inequality for any point $$P$$ inside the angle $$ACB$$;

3. Take $$P'$$ symmetric to $$P$$ respect to bisector of $$\angle ACB$$ and write an inequality 1. for the point $$P'$$.

• I couldn't really get part 3, but you can't guarantee that $P'$ is going to be inside the triangle... – aradarbel10 Jan 11 at 10:32
• @aradarbel10 Yes, of course, but $P'$ is placed inside the angle ACB and we can use 2. – Michael Rozenberg Jan 11 at 11:05

$$ad_a + bd_b + c_dc = 2\times Area$$

$$d_c + R_c$$ is greater that the shortest line from from $$C$$ to side $$c$$

$$(d_c + R_c)c\ge 2\times Area$$

$$(d_c + R_c)c \ge ad_a + bd_b + c_dc\\ cR_c \ge ad_a + bd_b$$

• You solved 1. in my hint. It's an easiest part of the proof. See please my post. – Michael Rozenberg Jan 11 at 6:41