Geometry

[obsidian patio]

Ive been struggling with this for a while

[obsidian patio]

In the parallelogram ABCD angle ADK = C is given
AF bisects the angle KAD
DE bisects the angle ADO
Prove: AF = DE

[shell shuttle]

you got this

[obsidian patio]

been studying for 14 hours a day lately

[ancient lava]

if af bisects the angle kad does that mean line AK = line AD??

[shell shuttle]

hi

[half bay]

What r u meant to find

[obsidian patio]

AF = DE

[half bay]

So are we supposed to prove that AF = DE

[blazing hull]

In the parallelogram ABCD angle ADK = C is given

Assuming that this means that the angles BCD and ADK are equal then this is pretty obvious since BCD = DAB.

And then, because DAB and ADK are equal, the whole trapezoid is symmetrical.

[half bay]

Ok thanks

[obsidian patio]

OH

how the hell did i not notice that

thanks.

[ancient lava]

DAB and ADK are not equal

[blazing hull]

Under the assumption that BCD = ADK, they are. If that's what OP meant.

[shell shuttle]

Yeah, cause DAB is the opposite angle in the parellelogram

so DAB=ADK=BCD

[fast grail]

.solved