Question 932426
BD is an altitude to AC, so *[tex \large \angle BDA = \angle BDC = 90^{\circ}]
Also *[tex \large \angle ABD = \angle DBC]
By looking at right triangles BAD and BCD, it follows that *[tex \large \angle BAC = \angle BCA] (note that these two triangles are furthermore congruent).
Therefore triangle ABC is isosceles because two of its interior angles (BAD = BAC, and BCA) are equal.