1957 AMC 12 Problem 44

Attempt Problem 44 of the 1957 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1957 AMC 12 solutions, or check the answer key.

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44.

In triangle

ABC,

AC=CD

and

\angle CAB-\angle ABC=30^\circ.

Then

\angle BAD

is:

30^\circ

20^\circ

22\frac12^\circ

10^\circ

15^\circ

Answer: E

Concepts:isosceles triangle angle chasing

Difficulty rating: 1630

Small hint:

Because

AC=CD,

triangle

ACD

is isosceles

Big hint:

Express

\angle CDA

using

\angle DAB

and

\angle ABC

Solution:

Let

\alpha=\angle CAB,

\beta=\angle ABC,

and

x=\angle BAD.

Since

D

lies on

BC,

\angle ADC=x+\beta

as an exterior angle of triangle

ABD.

Because

AC=CD,

triangle

ACD

is isosceles, so

\angle CAD=\angle ADC.

But

\angle CAD=\alpha-x.

Hence

\alpha-x=x+\beta,

so

2x=\alpha-\beta=30^\circ

and

x=15^\circ.

Thus, the correct answer is E.

Problem 44 in other years