Question

In the given figure, $PQR$ is a triangle and $S$ is any point in its interior.

Show that $PQ+PR>SQ+SR$.

Answer 1

Let $R^{\prime }$ and $r$ be the circum-radius of $\Delta$ PQR and $\Delta$ SQR.

Clearly, $R^{\prime }>r$.

Then using properties of triangle we have from $\Delta$ PQR

$PR=2R^{\prime }\sin \angle Q$ and $PQ=2R^{\prime }\sin \angle R$

Again from $\Delta$ SQR we get

$SR=2r\sin \angle SQR$ and $SQ=2r\sin \angle SRQ$.

Now, $PQ+PR=2R^{\prime }(\sin \angle R+\sin \angle Q)>2r(\sin \angle SQR+\sin \angle SRQ)=SR+SQ$ [Since $\angle R>\angle SRQ$ and $\angle Q>\angle SQR$]

So, $PQ+PR>SQ+SR$.