Russia
In ^@\triangle ABC^@, with ^@AB = AC^@, prove that the altitude from the vertex ^@A^@ bisects the side ^@BC^@.
Answer:
- We know that ^@\triangle 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. @^\implies AD\perp 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 ^@\triangle ADB^@ and ^@\triangle ADC,^@ we have@^ \begin{aligned} & AD = AD && [\text{Common}] \\ & AB = AC && [\text{Given}] \\ \therefore \space & \triangle ADB \cong \triangle ADC && [\text{By RHS criterion}] \end{aligned}@^
- As corresponding parts of congruent triangles are equal, we have@^BD = DC@^
- Thus, in an isosceles triangle, the altitude from the vertex bisects the base.
