geometrytrigonometry
Prove that the distance between the middle point of BC and the foot of the perpendicular from A is$\frac{b^2-c^2}{2a}$.
I found the length of $AM^2=\frac{2b^2+2c^2-a^2}{4}$ using Stewart theorem.I also found $AD^2$ using Stewart theorem but that is coming in terms of $\tan B$ and $\tan C$.If AD had come in terms of $a,b,c$,I would have solved the question using Pythagoras theorem.
Is my approach correct,if not how can i solve this question?Please help…..
In the picture shown, $AB = DC = 10$, $BC = AD = 6$, and $\angle BDC = 30^{\circ}$.
Apply the Law of Cosines to $\triangle BDC$. Then
$$6^2 = d^2 + 10^2 - 2 \cdot d \cdot 10 \cdot \cos 30^{\circ}$$ $$36 = d^2 + 100 - 20d \cdot \frac{\sqrt{3}}{2}$$ $$0 = d^2 - 10d\sqrt{3} + 64$$ $$d^2=10d\sqrt{3} + 64$$ $$d=\sqrt{10d\sqrt{3} + 64}$$
Apply the quadratic formula to solve for the long diagonal $d$. You should get two (geometrically valid) roots, although only one of the roots makes $BD$ the long diagonal; the other root makes it the short one.
Construct a cube with side length $\frac {\sqrt 2}{2}$
choose one vertex. There are 3 vertexes that are diagonally across each face.
Join these 4 vertexes.
You have constructed a regular tetrahedron with side length equal to 1.
See figure.
I have marked the centroid of two faces with the little dots.
Now lets move to the top view.... The centroid is $\frac 23$ the distance from the vertex to the opposite edge. Which means that the distance between the centroid of two faces is $\frac 13$ the lenght of the diagonal of the face. And we have already discussed that that length is 1.
$\frac 13$
Best Answer
$$c^2=BH^2+AH^2 \to c^2=(\frac{a}{2}-HM)^2+AH^2 $$ $$b^2=CH^2+AH^2 \to b^2=(\frac{a}{2}+HM)^2+AH^2 $$find $b^2-c^2$ $$b^2-c^2= (\frac{a}{2}+HM)^2+AH^2 -((\frac{a}{2}-HM)^2+AH^2 )=\\\frac{a^2}{4}+2\frac{a}{2}.HM +HM^2 +AH^2 -(\frac{a^2}{4}-2\frac{a}{2}.HM +HM^2 +AH^2)=2a.HM \\ \to \\HM=\frac{b^2-c^2}{2a}$$