Home
Class 12
MATHS
BE and CF are two equal altitudes of a t...

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isoscles.

Promotional Banner

Topper's Solved these Questions

  • TRIANGLES

    SUBHASH PUBLICATION|Exercise Exercise 5.4|3 Videos

  • TRIANGLES

    SUBHASH PUBLICATION|Exercise Exercise 5.5|3 Videos

  • TRIANGLES

    SUBHASH PUBLICATION|Exercise Exercise 5.2|7 Videos

  • THREE DIMENSIONAL GEOMETRY

    SUBHASH PUBLICATION|Exercise TRY YOURSELF|11 Videos

  • VECTOR ALGEBRA

    SUBHASH PUBLICATION|Exercise TRY YOURSELF|4 Videos

Similar Questions

Explore conceptually related problems

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

BE and CF are two altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

If a sin a=b sin B then prove that Delta ABC is isosceles.

If a sin a=b sin B then prove that Delta ABC is isosceles.

If in a triangle ABC, 2cosA=sinBcosecC , then :

In any triangle ABC , sin.(A)/(2) is

AX and DY are altitudes of two similar triangles Delta ABC and Delta DEF . Prove that AX : DY = AB : DE.

In a triangle ABC, if acosA=bcosB , then the triangle is

In Fig , altitudes AD and CE of triangle ABC intersect each other at the point P. show that : triangleAEP ~triangleCDP