1. How the angle bisector of a parallelogram form a rectangle?

2. If an angle of parallelogram is two-third of its adjacent angle. find the angles of the parallelogram?

3. Find the measures of all the angles of a parallelogram is one angle is 24 degree less than twice the smallest angle?

4. AB and CD are two parallel lines and a perpendicular angle intersect AB at X and CD at Y. Prove that the bisector of the interior angle form a rectangle?

5. ABCD is a parallelogram and line segment AX bisects the angle A and C respectively, show that AX II CY.

6. Given ABC, lines are drawn through A, B, and C respectively parallel to the side BC, CA and AB forming triangle PQR. Show that BC = 1/2 QR.

7. BM and CN are perpendicular to a line passing through the vertex A of a triangle ABC, if the angle is the midpoint of BC. Prove that angle M = angle N.

**(1)**

Hence, PQRS is a rectangle.

**(2)**

**(3)**

Let the smallest angle be = x^{0}

As we know that in a II^{gm} opposite angles are equal

Therefore, there will be two angles with x^{0} each.

Since it is given that one angel is 24 degree less than twice the smallest angle, which gives 2x-24

Sum of all angles of parallelogram is =360

Therefore, we get

2x-24+2x-24+x+x=360

6x-48=360

6x=360+48

6x=408

x=408/6=68

Therefore, the smallest angle is of 68^{0} .

And other angle = 2x-24=2 x 68-24=112

Hence angles of the parallelogram are : 112, 112, 68, 68.

**(4)**

AB and CD are two parallel lines intersected by a transversal. X and Y are the points of intersection *l *with AB and CD respectively. XP, XQ, YP and YQ are the bisector of ∠AXY, ∠BXY, ∠CYX and ∠DYX respectively.

AB || CD and *l* is the transversal,

∴ ∠AXY = ∠DYX (Pair of alternate angles)

∠AXY = ∠DYX

⇒ ∠1 = ∠4 (∠1 = ∠AXY and ∠4 = ∠DYX)

⇒ PX || YQ ...(1) **(If a transversal intersect two lines in such a way that pair of alternate interior angles are equal, then the two lines are parallel**)

Also, ∠BXY = ∠CYS (Pair of alternative angles)

∠BXY = ∠CYX

⇒ ∠2 = ∠3 (∠2 = ∠BXY and ∠3 = ∠CYX)

⇒ PY || XQ ...(2) **(If a transversal intersect two lines in such a way that pair of alternate interior angles are equal, then the two lines are parallel**)

From (1) and (2), we get

PX || YQ and PY || XQ

Hence, PXQY is a parallelogram ...(3) (**A quadrilateral is parallelogram, if both pair of its opposite sides are parallel**)

Now, ∠CYD = 180°

∠CYD = × 180° = 90°

(∠CYX + ∠DYX) = 90°

∠CYX + ∠DYX = 90°

⇒ ∠3 + ∠4 = 90°

⇒ ∠PYQ = 90° ...(4)

Thus, PXQY is a rectangle (Using (3) and (4))

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