The internal bisectors of the angles of a triangle ABC meet the sides in D,E,and F.Show that area of

anginih86

anginih86

Answered question

2022-06-15

The internal bisectors of the angles of a triangle ABC meet the sides in D,E,and F.Show that area of the triangle DEF is equal to 2 Δ × a b c ( b + c ) ( c + a ) ( a + b ) ,here Δ is area of triangle ABC

If I choose B as origin,C as(a,0),a is side length BC.what should i take coordinates of A,can i get answer with this approach or any better approach?

Answer & Explanation

jmibanezla

jmibanezla

Beginner2022-06-16Added 17 answers

Let D , E , F be on B C , C A , A B respectively.

Now, since we know that
B D : D C = c : b , A F : F B = b : a
the area of B D F is
Δ × c b + c × a a + b = a c ( a + b ) ( b + c ) Δ .
Similarly, we can have
[ C E D ] = a b ( c + a ) ( b + c ) Δ , [ A E F ] = b c ( a + b ) ( c + a ) Δ .
Thus, the area of D E F is
[ A B C ] [ B D F ] [ C E D ] [ A E F ] = 2 a b c ( a + b ) ( b + c ) ( c + a ) Δ .
hawatajwizp

hawatajwizp

Beginner2022-06-17Added 10 answers

If D B C, E C A, and F A B, then A B B C = A D D C . We have similar conditions for E and F. Then, calculate [ A E F ] [ A B C ] , [ B F D ] [ A B C ] , and [ C D E ] [ A B C ] , where [ P ] is the area of a geometric object P P.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Elementary geometry

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?