The height, the median line and the bisector of the vertex of the isosceles triangle coincide.

It is known that the triangle ABC is an isosceles triangle and the AD is the midline.

It is proved that AD bisects BC vertically and BD=DC isosceles triangle ABC(AB=AC).

Triangle ABC is an isosceles triangle (known)

∴AB=AC (the nature of isosceles triangle)

∴∠b =∞∠c (equilateral and equiangular)

∫AD is the center line (known)

∴BD=DC (the center line of isosceles triangle is perpendicular bisector)

Ad is the male edge.

∴△ADB≌△ADC(S.A.S)

Available ∠BAD=∠CAD, ∠ADB=∠ADC (congruent triangles corresponding angles are equal).

∠∠ADB+∠ADC =∠BDC (proven), and ∠BDC= 180 degrees (right angle definition).

∴∠ADB=∠ADC=90 degrees, and AD is perpendicular to BC.

(1) If the bisector of any angle of a triangle coincides with the median line of the opposite side, the triangle is an isosceles triangle. A triangle is an isosceles triangle if the bisector of any angle of the triangle coincides with the height of the opposite side. If the midline of any side of a triangle coincides with the height of that side, then this triangle is an isosceles triangle. To sum up: in a triangle, if the high line of one side coincides with the middle line of that side and coincides with any two lines in the bisector of the opposite corner of that side, it can be inferred that the triangle is an isosceles triangle.