Triangles ABC and ABD are isosceles with `AB = AC = BD, and bar(BD)` intersects `bar(AC)` at E. If `bar(BD) _|_ bar(AC)` then `/_C+ /_D` is

Triangles ABC and ABD are isosceles with AB=AC=BD, and bar(BD) intersects bar(AC) at E.If bar(BD)perpbar(AC) then /_C+/_D is

ABC is an isosceles triangle AB=AC and BD and CE are its two medians. Show that BD=CE.

Consider all triangels ABC satisfying the following conditions : AB = AC, D is a point on bar(AC) for which bar(BD) _|_ bar(AC), AD, and CD are integers, and BD^2 = 57. Among all such triangles, the smallest possible value of AC is

ABC is an isosceles triangle with AB =AC and BD,CE are its two medians. Show that BD=CE .

ABC is an isosceles triangle with AB =AC and BD,CE are its two medians. Show that BD=CE .

In /_\ABC , bar(AD) and bar(BE) are the midpoints of bar(BC) and bar(AC) . If AC=10 cm , then CE =

ABC is an isoscles triangle with bar(AB)= bar(AC) and bar(AD) is one of its altitudes. Is BD = CD ? Why or why not?

In an isosceles right triangle ABC, angleB=90^(@) and bar(BD) bot bar(AC) .Then BD=……………… ((AB)/2//(BC)/2 // (AC)/2) .