1. ## Triangle related question

If D is the mid-point of CA in triangle ABC and $\displaystyle \Delta$ is the area of triangle,then show that
$\displaystyle \tan(\angle ADB) = \frac{4\Delta}{a^2 - c^2}$
where a=BC,b=CA and c=AB

2. Hello,
Originally Posted by pankaj
If D is the mid-point of CA in triangle ABC and $\displaystyle \Delta$ is the area of triangle,then show that
$\displaystyle \tan(\angle ADB) = \frac{4\Delta}{a^2 - c^2}$
where a=BC,b=CA and c=AB
Please have a look at the sketch.

BE is the altitude of the triangle coming from B.
Thus $\displaystyle \mathcal{A}_{\text{ABC}}=\Delta=\frac{\text{BE} \times b}{2} \implies \text{BE}=\frac{2 \Delta}{b}$

Note that you want $\displaystyle \tan(\angle \text{ADB})$. Obviously, it is equal to $\displaystyle \frac{\text{BE}}{\text{ED}}=\boxed{\frac{2 \Delta}{b \times \text{ED}}}$ because the triangle BED is a right angle triangle in E.

Now, you'll have to calculate ED, since you already know the others.

D is the midpoint of AC. Therefore $\displaystyle \text{AD}=\frac{\text{AC}}{2}=\frac b2$
And $\displaystyle \text{ED}=\text{AD}-\text{AE}=\frac b2-\text{AE}$

If we multiply by 2, we can write $\displaystyle 2\text{ED}=(b-\text{AE})-\text{AE}=\text{CE}-\text{AE}$
Applying Pythagoras theorem in AEB, we have $\displaystyle \text{AE}=\sqrt{c^2-\text{BE}^2}$
Applying Pythagoras theorem in BEC, we have $\displaystyle \text{CE}=\sqrt{a^2-\text{BE}^2}$
$\displaystyle \implies \boxed{\text{ED}=\frac{\sqrt{a^2-\text{BE}^2}-\sqrt{c^2-\text{BE}^2}}{2}}$

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See the two boxed formula ?

This gives :

$\displaystyle \tan(\angle \text{ADB})=\frac{4 \Delta}{b \times \left(\sqrt{a^2-\text{BE}^2}-\sqrt{c^2-\text{BE}^2}\right)}$

Multiply by $\displaystyle \frac{\sqrt{a^2-\text{BE}^2}+\sqrt{c^2-\text{BE}^2}}{\sqrt{a^2-\text{BE}^2}+\sqrt{c^2-\text{BE}^2}}=\frac{\text{AE}+\text{EC}}{\sqrt{a^2-\text{BE}^2}+\sqrt{c^2-\text{BE}^2}}=\frac{b}{\sqrt{a^2-\text{BE}^2}+\sqrt{c^2-\text{BE}^2}}$

(in fact, the important step was to multiply the denominator by its conjugate, and it just popped up that it was the length b)
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There may be a shorter way, but I don't know... Inspire yourself from it

3. Hello Moo,I am really inspired by your diagram.Here we go:
$\displaystyle cos \theta=\frac{BD^2 + \frac{b^2}{4} - c^2}{2.BD.\frac{b}{2}}=\frac{\frac{2a^2 + 2c^2 - b^2}{4} + \frac{b^2}{4} - c^2}{b.BD}=\frac{a^2 - c^2}{2.b.BD}$

Now,using sine rule,we have
$\displaystyle \frac{\sin \theta}{c} = \frac{\sin A}{BD}$ .
$\displaystyle BD=\frac{c\sin A}{\sin \theta}=\frac{ac}{2R\sin \theta}$

Therefore,eliminating BD we get
$\displaystyle \tan \theta=\frac{abc}{(a^2 - c^2)R}=4\frac{abc}{4R}.\frac{1}{a^2 - c^2}=\frac{4\Delta}{a^2-c^2}$
I would want to hear more about that dyslexic and someone plz solve the other trigonometry question which i posted yesterday