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Question 1011972: In ΔABC, AB = CB and D is on AB such that AC = DC. If m∠ADC = 75, what is m∠B?
Image: https://gyazo.com/c2dac4c89fdb464ce1ccad5fc475a3a8
Sorry if its messy, but I attempted it and got stuck. My work so far was that if CA is congruent to CD then the triangle is isosceles so therefore ∠CAD is congruent to ∠CDA. Because ∠CDA was given as 75 degrees in the given I also know ∠CAD is 75 degrees. Also because ∠CDA and ∠CDB form a linear pair, they are also supplementary so I subtract 75 (∠CDB) from 180 to get ∠CDB which is 105. As I know the interior angles add up to the exterior angle I did 105 - 75 which gives me ∠ACD which is 30 degrees.
This is where I get stuck and have no idea how to find ∠B. I do not think I can use the ext. angle thm. as I do not know ∠DCB. Please help me out!
Answer by macston(5194)   (Show Source):
You can put this solution on YOUR website! .
Triangle ABC is isosceles, with angle CAB=angle ACB.
Triangle ADC is isosceles with angle CAD=angle CDA.
Given angle CDA=75 degrees, angles CAB (∠A) and ACB (∠C) also equal 75 degrees.
Angles CAB (∠A), and CBA (∠B), ACB (∠C) form the triangle ABC,
thus their sum is 180 degrees:
∠A + ∠B + ∠C = 180 degrees
75 degrees + ∠B + 75 degrees = 180 degrees
∠B = 30 degrees
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ANSWER The measure of Angle B is 30 degrees.
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