ΔABC and ΔDBC lie on the same side of BC, as shown in the figure. From a point, P on BC, PQ || AB and PR || BD are drawn, meeting AC at Q and CD at R respectively. Prove that QR || AD.
We can observe two triangles in the figure.
In ∆ABC,
PQ ∥ AB
Applying Thale’s theorem, we get
…(i)
In ∆BDC,
PR ∥ BP
Applying Thale’s theorem, we get
…(ii)
Comparing equations (i) and (ii),
Now, applying converse of Thale’s theorem, we get
QR ∥ AD
Hence, QR is parallel to the AD.