The base of an isosceles triangle is 2 root 2 if the medians to the equal sides intersect at right angles then the height of the triangle in

Hans007 Oct 28, 2017

#1**+2 **

Look at the following :

I think you want to find the height of the triangle containing the isosceles triangle

AB = 2sqrt (2)

Let AEB be isosceles......with angles ABE, BAE equal....so AE = BE

Bisect AEB.......by SAS......BH = AH = EH

And ABE = 90......so ABE, BAE = 45

And BH = sqrt (2).......so EH = sqrt (2)

Let E be the intersection of medians BG and AF

Draw GF through D.....drop perpendiculars from F, G to interesect AB at I and J and by AAS, triangle FIB is congruent to triangle GJA. ...then FI = GJ ...so.. GF is parallel to AB

And triangle AGF is similar to triangle ABC.....so DF = 1/2 of HB = sqrt(2)/2

So CD = 1/2 CH

And EDF is right and angle DFE = BAE = 45

So DEF = 45...so triangle EDF is 45 - 45 - 90 with DF = DE = sqrt (2) / 2

So.....HD = EH + DE = sqrt (2) + sqrt (2) / 2

And..........By similar triangles...... 2 CD = 2 HD = HC

So .....HC = 2 [ sqrt (2) + sqrt (2) / 2 ] = 3sqrt (2) = height of ABC

CPhill Oct 29, 2017