Question

21. In the given figure, ABCD is a parallelogram. X is the midpoint of AB. If produced CX intersects produced DA at Y, then prove that AYBC is a parallelogram.
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Answer 1

Step-by-step explanation:

so now ABCD is a parallelogram so ZA = ZC= y deg (opp angles of ||gm are equal)

so ZCBA = 180 - y (as adjacent angles of ||gm sum up to 180)

and AC is a diagonal and diagonals of || gm

and it bisects both ZA and ZC

so ZCAX = y/2

now AD || BC

and AD produces to Y

so AY || BC

ZCBA= ZXAY= 180 - y

similarly

ZCAX = ZXBY = y/2

now in triangle ABC and in triangle BAY

ZXAY = ZCBX

ZCAX = ZXBY

AX = XB

so both of triangles are congruent

now

AYB = ACB

AC = BY

AY = BC

by CPCT

now

AYBC Follows all the properties of parallelogram

• opposite angle are equal

• opposite sides are equal and parallel

• Diagonal bisects each other

• sum of angles are 360 deg .

hence proved

hope it'll help you.