Question

In a rhombus ABCD,EABF is a straight line in which EA=AB=BF, ED and FC produce they are meet at point X then prove that angleEXF is a right angle
Answer

In a rhombus all sides are equal
EA=AB (given),BF=AB(given)
Then EA=AD(AD=AB)
Similarly BF=BC(AB=BC)
angleAEX=angleADE(equal side opposite to equal angle)
Similarly angleBFX=angleBCF(equal side opposite to equal angle)

Let angleAEX=x and angleBFX=y
angle dab=2x (exterior angle property)
angle CBA=2y (exterior angle property)

2x+2y=180 degree
x+y=90 degree
angleE+ angleF= 90 degree
In triangle EXF
angle E + angle F + angle X=180
Then ,angle AXF=90 degree

HENCE PROVED.

Answer 1

What's this?? But Nice...