# Interesting problem in congruence of triangles

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.

• What is this dot "." in AB=AC.AD? And what is D? – zoli Jul 25 '15 at 12:08
• It is a dot to mark the end of the sentence. – wythagoras Jul 25 '15 at 12:10
• The dot is known as a full stop. – peterwhy Jul 25 '15 at 12:10
• @zoli-I have edited the question. – SHM Jul 25 '15 at 12:16

The proof is correct.

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

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

• @wythagoras-∠ABC is the same as ∠ABD.And more important he is comparing the triangles ABD and ACD.So he cannot write ∠ABC.He must write ∠ABD which is the part of triangle ABD. – SHM Jul 25 '15 at 12:20
• @soham this change makes the proof clearer, but I personally won't classify this as a logical defect. – peterwhy Jul 25 '15 at 12:24
• @peterwhy Somehow I had taken $AD$ to be parallel, rather than perpendicular. But I think the proof is correct. – wythagoras Jul 25 '15 at 12:42

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 $\angle BAD=\angle CAD$.

• This is not even the same logic. OP ('s student) used AAS with corresponding sides $AB=AC$ to prove congruence, but you did not even compare the right pair of corresponding sides. – peterwhy Jul 25 '15 at 20:42
• @James Pak-The student did use corresponding sides and angles.What is wrong with it? – SHM Jul 26 '15 at 5:55
• @peterwhy, soham: You both are right and I distorted the AAS reason. Here is an edit. Read and comment. – James Pak Jul 26 '15 at 6:37