Problem:
Quadrilateral ABCD has AB=BC=CD,∠ABC=70∘, and ∠BCD=170∘. What is the degree measure of ∠BAD?
Answer Choices:
A. 75
B. 80
C. 85
D. 90
E. 95
Solution:
Let M be on the same side of line BC as A, such that △BMC is equilateral. Then △ABM and △MCD are isosceles with ∠ABM=10∘ and ∠MCD= 110∘. Hence ∠AMB=85∘ and ∠CMD=35∘. Therefore
∠AMD​=360∘−∠AMB−∠BMC−∠CMD=360∘−85∘−60∘−35∘=180∘.​
It follows that M lies on AD and ∠BAD=∠BAM=(C)85∘​.
OR
Let â–³ABO be equilateral as shown.
Then
∠OBC=∠ABC−∠ABO=70∘−60∘=10∘.
Because ∠BCD=170∘ and OB=BC=CD, the quadrilateral BCDO is a parallelogram. Thus
OD=BC=AO and △AOD is isosceles. Let α=∠ODA=∠OAD. The sum of the interior angles of ABCD is 360∘, so we have
360=(α+60)+70+170+(α+10) and α=25.
Thus ∠DAB=60+α=(C)85∘​.
The problems on this page are the property of the MAA's American Mathematics Competitions
.jpg)