Question

if the median a.m. the angular bisector aidi and the altitude aidi drawn from vertex a of triangle ABC the wide angle into 4 equal parts and D lies between H and M then ?​

Answer 1

Answer:

Let Angle ACH =Angle HAK =Angle KAM =Angle MAB = x°

Since Angle AHC = 90°, in triangle ACH, Angle C = (90 - x)°

In triangle AHM, ext. angle AHC = 90°

= Angle HAM + Angle AMH = 2x + Angle AMH =90°

Therefore, Angle AMH = (90 - 2x )° ———————————(1)

From M draw MV parallel to AB.

Angle AMV = 3x° ———-(alt. angles)

Angle AMB and Angle AMC being linear pair, using (1),We get

Angle AMB= 180° - Angle AMC =180° - Angle AMH =( 180 - ( 90 -2x))° =(90 +2x)°

Angle AMV = Angle AMB - Angle VMB =( (90 + 2x ) - (90 - x))ï =3x°

In triangle AMH, ext. Angle AMB = (90 +2x)

AngleBMV = Angle AMB - Angle AMV =(90 + 2x) - 3x = (90 -x)

In triangle AMV,Angle AVM + Angle AMV + Angle MAV = 180

Angle AVM +3x +x =180