If two parallel lines are cut by a transversal , prove that the bisector of the interior amiles on the same side of the transversal intersect each other at right angle
Answers
Given-,EG , FG, EH and FH are bisectors of the interior angles ∠AEH, ∠CFE, ∠BEF and ∠EFD respectively..
To prove :- EFGH is a rectangle.
PROOF:- AB || CD and a transversal t cuts them at E anf F respectively .
:. ∠AEF=∠EFD [ alternate interior ∠s]
= 1/2 ∠AEF=1/2∠EFD = ∠GEF=∠EFH
but , these are alternate interior angles formed when the transversal EF cuts EG and FH
:. EG || FH. simillarly , EH|| FG
:. EGFH is a parallelogram.
Now, ray EF stands on AB.
:. ∠AEF + ∠BEF = 180° (linear pair)
= 1/2∠AEF + 1/2∠BEF= 90° = ∠GEF + ∠ HEF= 90°
= ∠GEH = 90° [ ∠GEF + ∠ HEF= ∠GEH]
Thus, EFGH is a parallelogram , one of whose angles is 90°
:. EFGH is a rectangle..
Solutions:
We know that the sum of interior angles on the same side of the transversal is 180°.
Hence, ∠BMN + ∠DNM = 180°
=> 1/2∠BMN + 1/2∠DNM = 90°
=> ∠PMN + ∠PNM = 90°
=> ∠1 + ∠2 = 90° ............. (i)
In △PMN, we have
∠1 + ∠2 + ∠3 = 180° ......... (ii)
From (i) and (ii), we have
90° + ∠3 = 180°
=> ∠3 = 90°
=> PM and PN intersect at right angles.
