In △ABC, with AB=AC, prove that the altitude from the vertex A bisects the side BC.
Answer:
- We know that △ABC is an isosceles triangle in which AB=AC.
Let AD be the altitude from the vertex A on the side BC.
As the altitude from a vertex to the opposite side is perpendicular to the opposite side. ⟹AD⊥BC Let us now represent the above situation with the help of a figure.
- We need to prove that BD=DC.
- In the right-angled △ADB and △ADC, we haveAD=AD[Common]AB=AC[Given]∴ △ADB≅△ADC[By RHS criterion]
- As corresponding parts of congruent triangles are equal, we haveBD=DC
- Thus, in an isosceles triangle, the altitude from the vertex bisects the base.