In a triange ABC, AD is the altitude from A. Given `bgtc,/_C=23^0 and AD= (abc)/(b^2+c^2), then /_A…..`

In triangle A B C ,A D is the altitude from Adot If b > c ,/_C=23^0,a n dA D=(a b c)/(b^2)-c^2, then /_B=_____

In a DeltaABC , AD is the altitude from A. Given b gt c, angleC=23^(@)" and "AD=(abc)/((b^(2)-c^(2)) , find angleB .

In a triangle ABC, AB = AC. Show that the altitude AD is median also.

In an equilateral triangleABC, AD is the altitude drawn from A on the side BC. Prove that 3AB^(2) = 4AD^(2)

In a triangle ABC with altitude AD, angleBAC=45^(@), DB=3 and CD=2 . The area of the triangle ABC is :

In a triangle ABC with altitude AD , angleBAC=45^(@), DB=3 and CD=2 . The area of the triangle ABC is : (a) 6 (b) 15 (c) 15/4 (d) 12

In a scalene triangle ABC , AD and CF are the altitudes drawn from A and C on the sides BC and AB repectively.If the area of the triangle ABC and BDF are 18sq.units and 2 sq. units respectively and DF = 2sqrt2 , then R =

In a triangle ABC, B=90^@ and D is the mid-oint of BC then prove that AC^2=AD^2+3 CD^2

In an acute angled triangle ABC, AD is the median in it. Prove that : AD^(2) = (AB^(2))/2 + (AC^(2))/2 - (BC^(2))/4

Statement I In any triangle ABC, the square of the length of the bisector AD is bc(1-(a^(2))/((b+c)^(2))). Statement II In any triangle ABC length of bisector AD is (2bc)/((b+c))cos ((A)/(2)).