Example 4
Two parallel lines l and m are intersected by a transversal p. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.
Given: line l ∥ m
transversal p intersects l & m at A & C resp.
Bisector of ∠ PAC & ∠QCA meet at B
& Bisector of ∠ SAC & ∠RCA meet at D
To prove: ABCD is a rectangle
Proof: We know that a rectangle is a parallelogram with one angle 90°
First we will prove ABCD is a parallelogram,
For l ∥ m & transversal p
∠ PAC = ∠ ACR
So ,1/2 ∠ PAC = 1/2 ∠ ACR
i.e., ∠ BAC = ∠ ACD
For lines AB and DC with AC as transversal
∠ BAC & ∠ ACD are alternate angles,
and they are equal
So, AB ∥ DC
Similarly,
for lines BC & AD, with AC as transversal
∠ BAC & ∠ ACD are alternate angles, and they are equal
∴ BC ∥ AD
Now, In ABCD,
AB ∥ DC & BC ∥ AD
As both pairs of opposite sides are parallel,
ABCD is a parallelogram.
Also for line l ,
∠ PAC + ∠ CAS = 180°
Multiplying both sides by half
1/2 ∠ PAC + 1/2 ∠ CAS = 1/2 × 180°
∠ BAC + ∠ CAD = 90°
∠ BAD = 90°
So, ABCD is a parallelogram in which one angle is 90°.
∴ ABCD is a rectangle.

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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