Question

Using ruler and compasses only, construct a parallelogram ABCD using the following data :
AB = $6$ cm, AD = $3$ cm, and $\angle DAB={45}^{\circ }$. If the bisector of $\angle DAB$ meets DC at P, prove that $\angle APB$ is right angle.

Answer 1

Steps :
(i) Draw AB = 6 cm.
(ii) With A as a center draw a line AX such that $\angle BAX={45}^{\circ }$
(iii) With A as a center and radii, 3 cm draw an are on AD.
(iv) Now with D and B as a center and radii 6 cm and 3 cm draw arcs cutting each other at C.
(v) Join DC and BC.
ABCD is the required parallelogram.
Here
$\angle PAB=\angle APD$ [Altemate angles]
$\angle CPB=\angle PBA$ [Altemade angles]
Now,
$\angle DPB+\angle APD+\angle CPB={180}^{\circ }$.....(i)
Also, considering $\Delta APB,$
$\angle PAB+\angle PBA+\angle APB={180}^{\circ }$......(ii)
Therefore, from (i) and (ii)
$\angle APB={90}^{\circ }$
Hence proved.