Example 2 : In Fig. 6.10, ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of ∠ POS and ∠ SOQ, respectively. If ∠ POS = x, find ∠ ROT.
Solution – Ray OS stands on the line POQ. Therefore, ∠ POS + ∠ SOQ = 180° But, ∠ POS = x Therefore, x + ∠ SOQ = 180° So, ∠ SOQ = 180° – x Now, ray OR bisects ∠ POS, therefore
Example 3 : In Fig. 6.11, OP, OQ, OR and OS are four rays. Prove that ∠ POQ + ∠ QOR + ∠ SOR + ∠ POS = 360°.
Solution – In Fig. 6.11, you need to produce any of the rays OP, OQ, OR or OS backwards to a point. Let us produce ray OQ backwards to a point T so that TOQ is a line (see Fig. 6.12). Now, ray OP stands on line TOQ. Therefore, ∠ TOP + ∠ POQ = 180° (1) (Linear pair axiom) Similarly, ray OS stands on line TOQ. Therefore, ∠ TOS + ∠ SOQ = 180° (2) But ∠ SOQ = ∠ SOR + ∠ QOR So, (2) becomes ∠ TOS + ∠ SOR + ∠ QOR = 180° (3) Now, adding (1) and (3), you get ∠ TOP + ∠ POQ + ∠ TOS + ∠ SOR + ∠ QOR = 360° (4) But ∠ TOP + ∠ TOS = ∠ POS Therefore, (4) becomes ∠ POQ + ∠ QOR + ∠ SOR + ∠ POS = 360°
Example 4 : In Fig. 6.19, if PQ || RS, ∠ MXQ = 135° and ∠ MYR = 40°, find ∠ XMY. Fig. 6.19 Fig. 6.20
Solution – Here, we need to draw a line AB parallel to line PQ, through point M as shown in Fig. 6.20. Now, AB || PQ and PQ || RS. Therefore, AB || RS (Why?) Now, ∠ QXM + ∠ XMB = 180° (AB || PQ, Interior angles on the same side of the transversal XM) But ∠ QXM = 135° So, 135° + ∠ XMB = 180° Therefore, ∠ XMB = 45° (1) Now, ∠ BMY = ∠ MYR (AB || RS, Alternate angles) Therefore, ∠ BMY = 40° (2) Adding (1) and (2), you get ∠ XMB + ∠ BMY = 45° + 40° That is, ∠ XMY = 85
Example 5 : If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.
Solution – In Fig. 6.21, a transversal AD intersects two lines PQ and RS at points B and C respectively. Ray BE is the bisector of ∠ ABQ and ray CG is the bisector of ∠ BCS; and BE || CG. We are to prove that PQ || RS. It is given that ray BE is the bisector of ∠ ABQ. Therefore, ∠ ABE = 1/2 ∠ ABQ (1) Similarly, ray CG is the bisector of ∠ BCS. Therefore, ∠ BCG = 1/2 ∠ BCS (2) But BE || CG and AD is the transversal. Therefore, ∠ ABE = ∠ BCG (Corresponding angles axiom) (3) Substituting (1) and (2) in (3), you get 1/2 ∠ ABQ = 1/2 ∠ BCS That is, ∠ ABQ = ∠ BCS
But, they are the corresponding angles formed by transversal AD with PQ and RS;
and are equal.
Therefore, PQ || RS
(Converse of corresponding angles axiom)
Example 6 : In Fig. 6.22, AB || CD and CD || EF. Also EA ⊥ AB. If ∠ BEF = 55°, find the values of x, y and z.
Solution – y + 55° = 180° (Interior angles on the same side of the transversal ED) Therefore, y = 180º – 55º = 125º Again x = y (AB || CD, Corresponding angles axiom) Therefore x = 125º Now, since AB || CD and CD || EF, therefore, AB || EF. So, ∠ EAB + ∠ FEA = 180° (Interior angles on the same side of the transversal EA) Therefore, 90° + z + 55° = 180° Which gives z = 35°
