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In triangle ABC , if AB = AC and D is a point on BC. Prove that AB^2 - AD^2 = BD × CD. I tried using Pythagoras theorem but to no avail. I heard about one Appollonius Theorem that could be used here but dont know how to.

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  • $\begingroup$ If $D$ is an arbitrary point on $\overline{BC}$, I don't see how Appollonius is supposed to be relevant. $\endgroup$ – Ted Shifrin Aug 3 '19 at 16:24
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Let $M$ be the midpoint of $\overline{BC}$. Without loss of generality, assume that $D$ lies on $\overline{MC}$. Then, using the Pythagorean Theorem a couple of times, \begin{align*} AB^2-AD^2 &= AC^2-AD^2 = (MC^2+AM^2)-(AM^2+MD^2) \\ &= MC^2-MD^2 = (MC+MD)(MC-MD) \\ &= (BM+MD)(MC-MD) = (BD)(DC), \end{align*} as required.

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  • $\begingroup$ This is a really clean way to do it, in that $AB^2-AD^2=MC^2-MD^2$ gets you to a difference of squares whose factors have an obvious meaning. $\endgroup$ – Ian Aug 3 '19 at 17:23
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Let $\angle BDA=\delta$.

\begin{align} \triangle ABD:\quad |AB|^2&=|AD|^2+|BD|^2-2\,|BD|\,|AD|\cos\delta \tag{1}\label{1} ,\\ \triangle ADC:\quad |AC|^2 =|AB|^2&=|AD|^2+|CD|^2-2\,|CD|\,|AD|\cos(\pi-\delta) \\ &=|AD|^2+|CD|^2+2\,|CD|\,|AD|\cos\delta \tag{2}\label{2} , \end{align}

$\eqref{1}\cdot |CD|+\eqref{2}\cdot |BD|\Rightarrow$

\begin{align} |AB|^2\cdot(|BD|+|CD|)&= |AD|^2\cdot(|BD|+|CD|) +|CD|\cdot|BD|^2 +|BD|\cdot|CD|^2 \\ &= (|AD|^2+|BD|\cdot|CD|)\cdot(|BD|+|CD|) ,\\ |AB|^2-|AD|^2 &=|BD|\cdot|CD| . \end{align}

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  • $\begingroup$ This answer would be a lot more legible with just a simple statement about why you consider (1)*CD + (2)*BD, which is to say to cancel out the terms involving $\cos(\delta)$. $\endgroup$ – Ian Aug 5 '19 at 1:39
  • $\begingroup$ @Ian: Right, without this deep thought, it does not give a correct answer to the question, and costs nothing, thanks. $\endgroup$ – g.kov Aug 5 '19 at 2:17

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