Question

AB,CD are two parallel lines and a transversal l intersects AB at G and CD at H.Prove that the bisectors of the interior angles form a parallelogram,with all it's angles right angles

Answer 1

◆Ekansh Nimbalkar◆
hello friend here is your required answer

Answer 2

$\mathfrak{\huge{\blue{Answer}}}$

Given :-

AB║CD and transversal ll intersects AB at X and Cd at Y.

To be proof :-

PYQX is a rectangle

PROOF :

\bf \angle{DYX}=\angle{AXY}∠DYX=∠AXY [Alternate interior angles]

\bf \frac{1}{3}\angle{DYX}=\frac{1}{2}\angle{AXY}

3

1

∠DYX=

2

1

∠AXY

∴ ∠1 = ∠2

Now,

XY intersects PX and QY at X and Y respectively,

such that

∠1 = ∠2

∴ PX║QY

Similarly,

PY║QX

So,

PYQX is a parallelogram

Now,

∠BXY + ∠DYX = 180° [consecutive interior angles]

Or

\bf 2\angle{2}+2\angle{3}=180^o2∠2+2∠3=180

o

\bf \angle{2}+\angle{3}=90^o∠2+∠3=90

o

⇒ ∠1 +∠3 = 90° [ ∴ ∠2 = ∠1 ]

⇒ ∠QXP = 90°

∴ PYQX is a rectangle

Hence proved.

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