Interesting problem in congruence of triangles

Question

While solving the exercises of my book I came across this interesting problem:

$\triangle ABC$ is isosceles triangle with $AB=AC$. D is a point on base BC such that $AD$ perpendicular on $BC$. To prove that $\angle BAD=\angle CAD$ a student does as follows. Between $\triangle ABD$ and $\triangle ACD$,

1. $AB=AC$ (given)
2. $\angle B=\angle C$ (because $AB=AC$)
3. $\angle ADB=\angle ADC$ ($=90^\circ$).

Therefore $\triangle ABD\cong \triangle ACD$. So, $\angle BAD=\angle CAD$. Now what is the defect in these arguments?

Now what is the defect in this argument? Please try to solve it.

Answer 1

The proof is correct.

$\angle B = \angle C$ is very vague. He should have written $\angle ABC = \angle ACB$.

But we can see that $\angle ACD = \angle ACB = \angle ABC = \angle ABD$.

Answer 2

Here is a counterexample of the argument, showing that the argument is not generally valid:

1. $BC=BC$ (common side)

2. $\angle ACB = \angle BCD$ (common angle)

3. $\angle ABC=\angle BDC$ (= $90^\circ$)

By the same logic, $\triangle ABC \cong \triangle BDC$. But they aren't. They are only similar triangles.

The key to the argument's apparent validity is that there, indeed, are equal corresponding sides of the two triangles in the student's case, namely, $BD=CD$ and $AD=AD$. But in this case there aren't. And the argument itself shows seeming congruence whether or not there are equal corresponding sides.

If the student really wanted to prove the congruence between the two triangles, simply use Pythagoras theorem.

Edit: Yes, the comments below are right.

In addition, if defect does not mean invalidity but imperfection, like indirectness, then there is. In the student's argument step 2 and step 3 are enough to prove $\angle BAD=\angle CAD$.