Question

$\Delta ABC$ is an acute angled triangle CD be the altitude through C. If AB = 8, CD = 6. Find the distance between the midpoints of AD and BC.

Answer 1

Let $P$ & $R$ are the mid-point of $AD$ & $BC$ respectively.

Now draw $RQ$ perpendicular to $AB$.

Now, in $△CDB$ & $△RQB,$

$\angle D=\angle Q=90°$

$\angle B=\angle B$

$\therefore △CDB\sim △RQB\text{(by}AA\text{)}$

So, basic propertionality theorem, if $R$ is mid point of $BC$.

$\therefore Q$ is midpoint of $BD$.

We can say,

$⟹AP+DQ=PD+QB$

$⟹2(AP+DQ)=PD+Q{B}_{A}P+DQ$

$⟹2(AP+DQ)=AB$

$⟹AP+DQ=\frac{AB}{2}$

$⟹AP+DQ=\frac{8}{2}=4$

Also,

$QR=\frac{1}{2}CD=\frac{1}{2}\times 6=3$

$\therefore PR=\sqrt{{\left(QR\right)}^2+{\left(PQ\right)}^2}$ (By pythagoras theorem)

$=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}$

$=\sqrt{25}$

$=5$