Question

ABC is a triangle and D is the middle point of BC. If AD is perpendicular to AC, prove that $cosAcosC=\frac{2(c^2-a^2)}{3ac}.$

Answer 1

Median AD is perpendicular to AC. From B, drawn BE perpendicular to CA produced. Then by geometry, we know that A is the mid-point of CE, so that CA=AE=b and if AD = y then BE = 2y.

$cosC=\frac{b}{a/2}=\frac{2b}{a}$ ........(1)

$cos(\pi -A)=b/c$

$\therefore cosA=-b/c$ .............(2)

$\therefore cosAcosC=-2b^2/ac$ ......(3)

Now we have to eliminate $b^2$ from (3). Apply Pythagoras theorem on ${△}^{s}BEA$ and DAC.

$4y^2+b^2=c^2$ (In $△$ BEA)

and $y^2+b^2=a^2/4$ (In $△$ DAC)

or $4y^2+4b^2=a^2$.

Subtracting, we get $3b^2=a^2-c^2$

$\therefore -b^2=(c^2-a^2)/3$

Putting in (3), we get $cosAcosC=\frac{2(c^2-a^2)}{3ac}$