" In "Delta," on the side "BC,D" and "E" are two points such that "BD=DE=EC." Also "/_ADE=/_AED=alpha," then "

In A B C , on the side B C ,Da n dE are two points such that B D=D E=E Cdot Also /_A D E=/_A E D=alpha, then

In a triangle ABC, points D and E are taken on side BC such that BD=DE=EC. If angle ADE = angle AED=theta, then: (A)tan theta=3tan B(B)3tan theta=tan C

In a triangle ABC, points D and E are taken on side BC such that BD= DE= EC. If angle ADE = angle AED = theta , then: (A) tan theta = 3 tan B (B) 3 tan theta = tan C

In Fig.30D and E are to points on BC such that BD=DE=EC. Show that ar(ABD)=ar(ADE)=ar(AEC)

In a triangle ABC, points D and E are taken on side BC such that BD= DE= EC. If angle ADE = angle AED = theta , then: (A) tan theta = 3 tan B (B) 3 tan theta = tan C (C) (6tan theta )/( tan^2 theta -9)= tanA (D) angleB = angle C

In a triangle ABC, if a = 5, b=4 and c = 3, D and E are the points on BC such that BD = DE = EC. If DeltaDAE =theta, then tan theta =

In Figure,AD=AE and D and E are points on BC such that BD=EC .Prove that AB=AC

In triangleABC , D and E are two points on AB and AC such that DE || BC and AD : BD = 3 : 2, then DE : BC =