Need help proving this geometry problem.

Question

My friend asked me one question yesterday.It is as follows.

Let there be two triangles ABD and ACD.D is a point on base BC such that BD=CD(given).Also,clearly side AD is common.Now we know median bisects a triangle into two parts of equal area.Now,two sides of the triangle are equal and area is also same.Apparently this seems that the sides AB will be equal to BC (quite funny actually....).Now how do I mathematically disprove that AB is not equal to AC?

Thank you for any help.Explanation with a diagram, if possible ,is greatly and highly appreciated..:-)

Answer 1

if $AB = AC$ then by the rule of $S-S-S$ those two are equal triangle...so $\angle ABD = \angle ADC = 90^o$ ...so now if we just start with a triangle which is not a right angle trianlge you can get the contradiction

Answer 2

if $\angle ADC = x$, then $\angle ADB= 180 - x$ (Angles on a straight line)

Now let's assume $AB=AC$:

This would mean the following:

1. $AB=AC$ (Assumption), $BD=DC$ (Given), and $AD=AD$ (Same line).

2. Thus $\triangle ABD \equiv \triangle ACD$ (S, S, S)

3. Thus $\angle ADC = \angle ADB$ ($\triangle ABD \equiv \triangle ACD$)

4. BUT $\angle ADC$ does not necessarily equal $\angle ADB$ ($x$ does not necessarily equal $180 - x$)

5. This means that $AB$ does not neccesarily equal $AC$ (Proof by contradiction)

(Also known as reductio ad absurdum)