# Median=half of the base, find vertex angle

In a $\triangle ABC$, $AB=AC$. $BA$ is produced to $D$ in such a manner that $AC=AD$. Find $\angle BCD$

I let $\angle ACB$ as $x$ and therefore, $\angle DAC$ becomes $2x$, so, $\angle ACD=90^\circ-x$. So, $\angle BCD=90^\circ$

But I am wondering why does this angle have to be a right angle only? We are just, in a way, given that the median of $\triangle BCD$ is half of $BD$. So, why would it make the vertex angle a right angle?

Moreover, if I assume a circle around $\triangle BCD$, why does the side $BD$ have to be a diameter. Why can't it be just a normal chord which is being bisected? What's special about $AB=AC=AD$?

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Do you mean find angle DBC? because BCD can be anything. –  mniip Feb 25 '14 at 16:07
@mniip- not angle DBC, but angle BCD –  Ramit Feb 25 '14 at 16:19
@Ramit OK, I see. There are several approaches to this problem. Do you know about this: en.wikipedia.org/wiki/Thales'_theorem ? –  Sawarnik Feb 25 '14 at 16:22
@Sawarnik-yes, I am aware that angle subtended in a semi circle is 90 degrees –  Ramit Feb 25 '14 at 16:22
@Ramit Does my answer completely satisfy your doubts? –  Sawarnik Feb 26 '14 at 6:57

I see that you have proved that $\angle BCD=90^{\circ}$, so I may use this result. On your second question:

There is a result known as Thales' theorem, which states that if $BD$ in the diameter (with reference to the diagram), then $\angle BCD=90^{\circ}$.

However, when you ask that when we assume a circle around the $\triangle BCD$, why does the side $BD$ have to be a diameter? The reason is that Thales theorem has a converse, which states that if $\angle BCD=90^{\circ}$, which we have proved, then $BD$ in the diameter. Why this happens, is proved in the link I gave above.

Your first question, is I think unclear in what you are asking for. We are given that the median of $\triangle BCD$ is half of $BD$ is an important and significant fact, and we can establish by mathematical proofs that it is enough.

Edit:

I should elaborate more on the median of ''$\triangle BCD$ is half of $BD$ is an important and significant fact''. What happens is that the median being equal to half the base, brings many properties and constructions, into play, using which we can solve the problem.

Note that Apollonius' theorem, gives us an explicit relation between a median and the sides, so having one more relation can lead to some useful simplification. And so it happens, using the extra information we have here, the theorem reduces to Pythagoras Theorem. The point is, some connections which are seemingly pointless, can bring us many unexpected results. That is the beauty of mathematics. Hope it helps :)

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Apollonius theorem made my day. Thanks! –  Ramit Feb 26 '14 at 6:58
@Ramit You understand the proof given at Wikipedia using cosine rule, or you want a geometric proof? –  Sawarnik Feb 26 '14 at 7:02
Actually I just wnated to prove the tringle as right angled using the properties of median. Apollonius theorem helped me in that. So, I am satisfied. –  Ramit Feb 26 '14 at 7:04