In triangle ABC, AB = AC and D is the midpoint of BC. Prove that AD is perpendicular to BC.

Step-by-step solution

1. 1

   List what is given and what “perpendicular” will require. We are given $AB=AC$ (the triangle is isosceles with base $BC$) and $BD=DC$ (definition of midpoint). To prove $AD\perp BC$ we must show that one of the angles $\angle ADB$ or $\angle ADC$ equals $90^\circ$. The plan is to prove those two angles are equal and then use the fact that they sit on a straight line, which forces each to be a right angle. That is a much cleaner route than trying to compute an angle directly.

2. 2

   Compare triangles $ABD$ and $ACD$. Segment $AD$ splits the original triangle into two smaller triangles. Match their parts:

   $AB=AC\quad(\text{given})$

   $BD=DC\quad(D\text{ is the midpoint of }BC)$

   $AD=AD\quad(\text{common side})$

   Each pair matches, so this is three sides against three sides.

3. 3

   Conclude congruence by SSS. By the side-side-side criterion,

   $\triangle ABD\cong\triangle ACD$

   SSS is the right choice here because all three given facts are about lengths; no angle information was supplied, so SAS or ASA would require an extra unproved assumption.

4. 4

   Transfer the equality to the angles at $D$. Corresponding parts of congruent triangles are equal, and $\angle ADB$ in the first triangle corresponds to $\angle ADC$ in the second:

   $\angle ADB=\angle ADC$

5. 5

   Use the straight line $BDC$ to pin the value at $90^\circ$. Because $D$ lies between $B$ and $C$ on segment $BC$, the two angles form a linear pair and are supplementary:

   $\angle ADB+\angle ADC=180^\circ$

   Substituting the equality from the previous step, $2\angle ADB=180^\circ$, hence

   $\angle ADB=\angle ADC=90^\circ\quad\Longrightarrow\quad AD\perp BC$

   Both facts were needed: equal angles alone do not give $90^\circ$, and supplementary angles alone do not either. It is the combination that closes the proof.

6. 6

   Verify with a coordinate proof. Place the base symmetrically: $B=(-b,0)$, $C=(b,0)$, so $D=(0,0)$ is the midpoint. The condition $AB=AC$ forces $A$ to be equidistant from $B$ and $C$, i.e. $A=(0,h)$ for some $h\neq0$. Then $\vec{DA}=(0,h)$ and $\vec{DC}=(b,0)$, and their dot product is

   $\vec{DA}\cdot\vec{DC}=0\cdot b+h\cdot0=0$

   A zero dot product means the segments are perpendicular, confirming the synthetic proof. As a bonus, the same picture shows $AD$ is simultaneously the median, the altitude, the angle bisector from $A$, and the perpendicular bisector of $BC$ — all four coincide precisely when the triangle is isosceles.