BD is the disector of angle ABC. From a point P in BD, perpendiculars PE and PF are drawn to AB and BC respectively, prove that :
(i) Triangle BEP is conguent to triangle BFP (ii) PE=PF.

ABC is an equilateral triangle,and AD and BE are perpendiculars to BC and AC respectively. Prove that: 1.AD=BE 2.BD =CE[Delta ABD=Delta BCE.]

An isosceles triangle ABC is right angled at B.D is a point inside the triangle ABC. P and Q are the feet of the perpendiculars drawn from D on the sides AB and AC respectively of Delta ABC . If AP = a cm, AQ = b cm and angle BAD= 15^(@), sin 75^(@)=

Let ABC be an equilateral triangle and P be a point within the triangle.Perpendicular PD,PE and PF are drawn to three sides of the triangle,then the value of (AB+BC+CA)/(PD+PE+PF) is

In Delta ABC ,AD is a perpendicular drawn from A to the side BC. If AD^(2)=BD . CD , then prove that the triangle is a right-angled triangle.

I is the incentre of the Delta ABC . The perpendicular drawn from I to BC intersects BC at. P. Prove that AB-AC=BP-CP .

ABC is right triangle right angled at B.If BD is the length of the perpendicular drawn from B to AC then

The incircle of an isosceles triangle ABC in which AB=AC , touches the sides BC, CA, and AB at D, E, and F respectively. Prove that BD=DC .

The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively . Prove that BD = DC.