In Triangle` A B C` with fixed length of `B C ,` the internal bisector of angle `C` meets the side `A B a tD` and the circumcircle at `E` . The maximum value of `C D×D E` is `c^2` (b) `(c^2)/2` (c) `(c^2)/4` (d) none of these

In DeltaABC with fixed length of BC , the internal bisector of angle C meets the side AB at D and meet the circumcircle at E. The maximum value of CD xx DE is

In A B C , the bisector of the angle A meets the side BC at D andthe circumscribed circle at E. Prove that D E=(a^2secA/2)/(2(b+c))

If the bisector of the angel C of a triangle ABC cuts AB in D and the circumcircle E, prove that CE:DE=(a+b)^2:c^2

Let ABC be a triangle inscribed in a circle and let l_(a)=(m_(a))/(M_(a)), l_(b)=(m_(b))/(M_(b)), l_(c )=(m_(c ))/(M_(c )) where m_(a), m_(b), m_(c ) are the lengths of the angle bisectors of angles A, B and C respectively , internal to the triangle and M_(a), M_(b) and M_(c ) are the lengths of these internal angle bisectors extended until they meet the circumcircle. Q. The maximum value of the product (l_(a)l_(b)l_(c))xxcos^(2)((B-C)/(2)) xx cos^(2)(C-A)/(2)) xx cos^(2)((A-B)/(2)) is equal to :

In triangle A B C ,/_C=(2pi)/3 and C D is the internal angle bisector of /_C , meeting the side A Ba tD . If Length C D is 1, the H.M. of aa n db is equal to: 1 (b) 2 (c) 3 (d) 4

In a triangle A B C , the internal bisectors of /_B\ a n d\ /_C meet at P and the external bisectors of /_B\ a n d\ /_C meet at Q . Prove that /_B P C+\ /_B Q C=180^0

If in a triangle A B C , the altitude A M be the bisector of /_B A D ,w h e r eD is the mid point of side B C , then prove that (b^2-c^2)=a^2//2.

The internal bisectors of the angles of a triangle ABC meet the sides in D, E, and F. Show that the area of the triangle DEF is equal to (2/_\abc)/((b+c)(c+a)(a+b) , where /_\ is the area of ABC.

In A B C ,/_A=50^0,/_B=70^0 and bisector of /_C\ m e e t s\ A B\ a t\ Ddot find the angles of the triangle A D C and B D Cdot

In A B C , internal angle bisector of /_A meets side B C in D . DE_|_A D meets A C at E and A B at F. Then (a) A E is in H.P. of b and c (b) A D=(2b c)/(b+c)cosA/2 (c) E F=(4b c)/(b+c)sinA/2 (d) AEF is isosceles