Question

In triangle $ABC;$ $P$ is mid-point of $AB,$ $Q$ is mid-point of $AC$ and $D$ is any point in base $BC$. Use Intercept Theorem to show that $PQ$ bisects $AD.$

Answer 1

Given : $ABC$ is a triangle, $PQ||BC;AD$ is the median which cuts $PQ$ at$R$.

To prove : $AD$ bisects $PQ$ at $R$.

Proof : In $\Delta ABD;PR||BD$

$\underset{–}{AP}=\underset{–}{AR}\left(BPT\right)$

$PBRD$

In $\Delta ACD,RQ||DC$

$\therefore \underset{–}{AR}=\underset{–}{AQ}\left(BPT\right)$

$RDRD$

In $\Delta APR$ and $\Delta ABD$,

$\angle APR=\angle ABD$ (corresponding angles)

$\angle ARP=\angle ADB$ (corresponding angles)

$\therefore \Delta APR$ is similar to $\Delta ABD$ (AA similarity)

$\therefore \underset{–}{AP}=\underset{–}{AR}=\underset{–}{PR}$ (corresponding sides of similar triangles are proportinal) ...(i)

$ABADBD$

Similarly $\Delta ARQ$ is similar to $\Delta ADC$

$\underset{–}{AQ}=\underset{–}{AR}=\underset{–}{RQ}...\left(ii\right)$

$ACADDC$

According to equation (i) and (ii)

$\underset{–}{AR}=\underset{–}{PR}=\underset{–}{RQ}$

$ADBDDC$

but $BD=DC$ (given)

$\therefore PR=RQ$

or $AD$ bisects $PQ$ at $R$