R S Aggarwal and Veena Aggarwal Solutions for Exercise 4: Exercise 7D

In $\Delta ABC, D$ is the midpoint of $BC$ and $AE\perp BC$. If $AC>AB$, show that $A{B}^2=A{D}^2-BC.DE+\frac{1}{4}B{C}^2.$

In the given figure, $D$ is the midpoint of side $BC$ and $AE\perp BC$. If $BC=a$, $AC=b$, $AB=c$, $AD=p$ and $AE=h$, prove that
$b^2=p^2+ax+\frac{a^2}{x}$

In the given figure, $D$ is the midpoint of side $BC$ and $AE\perp BC$. If $BC=a$, $AC=b$, $AB=c$, $AD=p$ and $AE=h$, prove that
$c^2=p^2-ax+\frac{a^2}{x}$

In the given figure, $D$ is the midpoint of side $BC$ and $AE\perp BC$. If $BC=a$, $AC=b$, $AB=c$, $AD=p$ and $AE=h$, prove that
$b^2+c^2=2p^2+\frac{a^2}{2}$

In the given figure, $D$ is the midpoint of side $BC$ and $AE\perp BC$. If $BC=a$, $AC=b$, $AB=c$, $AD=p$ and $AE=h$, prove that
$b^2-c^2=2ax$

In a $△ABC$, $AD$ is a median and $AL\perp BC$.

Prove that
$A{B}^2=A{D}^2-BC·DL+{\left(\frac{BC}{2}\right)}^2$
 

In a $△ABC$, $AD$ is a median and $AL\perp BC$.

Prove that
$A{C}^2+A{B}^2=2.A{D}^2+\frac{1}{2}B{C}^2$

Naman is doing fly-fishing in a stream. The tip of his fishing rod is $1.8 \mathrm{m}$ above the surface of the water and the fly at the end of the string rests on the water $3.6 \mathrm{m}$ away from him and $2.4 \mathrm{m}$ from the point directly under the tip of the rod. Assuming that the string (from the top of his rod to the fly) is taut, how much string does he have out (see the adjoining figure)? If he pulls in the string at the rate of $5 \mathrm{cm}$ per second, what will be the horizontal distance of the fly from him after $12$ seconds.