# MATHS

posted by .

AX is the bisector of <BAC p is any point on AX. prove that the perpendicular drawn from Pto AB and AC are equal

• MATHS -

let pb and pc be the perpendiculars to AB and AC.

Since Ax is the angle bisector, angle XAB = XAC

using similar triangles,

pb/pa = pc/pa
so,
pb=pc

## Similar Questions

1. ### Geometry

Can I get help starting this exercise:- Let triangle ABC be such that AB is not congruent to AC. Let D be the point of intersection of the bisector of angle A and the perpendicular bisector of side BC. Let E, F, and G be the feet of …
2. ### maths

In triangle ABC angle B > angle C if AM is the bisector of angle BAC and AN perpendicular BC. prove that angle MAN =1/2(angle B - angle C)
3. ### Maths

AX is the bisector of <BAC p is any point on AX. prove that the perpendicular drawn from Pto AB and AC are equal
4. ### math

in triangle abc ad is bisector of angle a and angle b is twice of angle c prove that angle bac is equal to 72
5. ### maths

in triangle abc angle b = 2angle c.d is a point ob bc such that ad bisects angle bac and ab = cd.prove that angle bac = 72
6. ### Geometry

BC is a triangle with ∠BAC=60∘,AB=5 and AC=25. D is a point on the internal angle bisector of ∠BAC such that BD=DC. What is AD^2?
7. ### maths

ABC is a triangle with ∠BAC=60∘,AB=5 and AC=25. D is a point on the internal angle bisector of ∠BAC such that BD=DC. What is AD^2?
8. ### Math

Two conjugate diameters of the ellipse x^2/a^2+y^2/b^2=1 are drawn to meet the directrix at P and Q. PM is drawn perpendicular to OQ and ON perpendicular to OP. Prove that PM and ON meet at a fixed point and state its coordinates.
9. ### Math

Please help me solve this question. Two conjugate diameters of the ellipse x^2/a^2+y^2/b^2=1 are drawn to meet the directrix at P and Q. PM is drawn perpendicular to OQ and ON perpendicular to OP. Prove that PM and ON meet at a fixed …
10. ### maths

BD is the bisector of angle ABC . prove that the perpendicular distances from BD to AB and CB are equal.

More Similar Questions

Post a New Question