Help! In quadrilateral $ABCD$, angle $BAD$ and angle $CDA$ are trisected as shown. What is the degree measure of angle $AFD$?

Question

Help! I have a math problem that's been confusing me a lot.

Source: AoPS Staff

The problem is:

In quadrilateral $ABCD$, angle $BAD$ and angle $CDA$ are trisected as shown. What >is the degree measure of angle $AFD$? This is an image that comes with the problem: https://i.stack.imgur.com/gsVKr.png

I will appreciate a hint or a full solution please

https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question — cqfd, Sep 02 '19 at 17:09

score 1 · Accepted Answer · answered Sep 02 '19 at 17:20

1

Sum of internal angles of a quadrilateral is $360°$ so you have $3x+3y+210°=360°\iff x+y=50°$. Sum of internal angles of a triangle is $180º$ so $AFD=180º-2x-2y=80°$

answered Sep 02 '19 at 17:20

Alessandro Cigna

2,496

Help! In quadrilateral $ABCD$, angle $BAD$ and angle $CDA$ are trisected as shown. What is the degree measure of angle $AFD$?

1 Answers1