P is any point in the angle ABC such that the perpendicular drawn from P on AB and BC are equal. Prove that BP bisects angle ABC.

Triangle ABC is isosceles with AB=AC and BC=65cm. P is a point on BC such that the perpendiculardistances from P to AB and AC are 24cm and 36cm, respectively. The area of triangle ABC (in sq cm is)

If AP bisects angle BAC and M is any point on AP, prove that the perpendiculars drawn from M to AB and AC are equal.

The vertices of a triangle ABC are the points (0, b), (-a, 0), (a, 0) . Find the locus of a point P which moves inside the triangle such that the product of perpendiculars from P to AB and AC is equal to the square of the perpendicular to BC .

ABCD is a parallelogram E is mid-point of AB and DE bisects angle D. Prove that CE bisects angle C

P is any point inside the triangle ABC. Prove that : angleBPC gt angleBAC .

In the following figure, ABCD is a parallelogram. Prove that: BP bisects angle B

In parallelogram ABCD, the bisector of angle A meets DC at P and AB= 2AD. Prove that: BP bisects angle B

A triangle ABC has angle B = angle C . Prove that : (i) the perpendiculars from the mid-point of BC to AB and AC are equal. (ii) the perpendicular from B and C to the opposite sides are equal.

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 a right traingle, right angled at C. if P is the length of perpendicular from C to AB and AB=c, BC=a and CA=b, then prove that (i) pc=ab (ii) 1/(p^(2)) = 1/(a^(2))+1/b^(2)