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.
Hence, PQRS is a rectangle.
Let the smallest angle be = x0
As we know that in a IIgm opposite angles are equal
Therefore, there will be two angles with x0 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
Therefore, the smallest angle is of 680 .
And other angle = 2x-24=2 x 68-24=112
Hence angles of the parallelogram are : 112, 112, 68, 68.
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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