Show that if there is a transversal between two parallel lines, the interior angle bisectors meet to form a rectangle. [4 MARKS]
Diagram : 1 Mark
Concept : 1 Mark
Proof : 2 Marks
Consider 2 parallel lines l and m.
It is given that PS∥QR and a transversal intersect them at points A and C respectively.
The bisector of ∠PAC and ∠ACQ intersect at B and bisectors of ∠ACR and ∠SAC intersect at D.
We are to show that quadrilateral ABCD is a rectangle.
Now, ∠PAC=∠ACR [Pair of alternate angles]
These form a pair of alternate angles for lines AB and DC with AC as transversal and they are equal also.
Similarly, BC|| AD.......(ii) [Considering ∠ACB and ∠CAD]
∴ quadrilateral ABCD is a parallelogram.
Also, ∠PAC+∠CAS=180∘ [linear pair]
So, ABCD is a parallelogram in which one angle is 90∘.
∴ ABCD is a rectangle.