Question

If \orange{\angle ADC}∠ADCstart color #ffa500, angle, A, D, C, end color #ffa500 measures 23^\circ23 ∘ 23, degrees, what does \blue{\angle ABC}∠ABCstart color #6495ed, angle, A, B, C, end color #6495ed measure?

Answer 1

Can you just post the pic

Answer 2

Answer:

23 degrees

Step-by-step explanation:

First add a diameter. The triangle \pink{ABD}ABDstart color #ff00af, A, B, D, end color #ff00af is isosceles as \pink{AB}ABstart color #ff00af, A, B, end color #ff00af and \pink{BD}BDstart color #ff00af, B, D, end color #ff00af are radii, so must be equal. Therefore, \pink{\angle{BAD}}∠BADstart color #ff00af, angle, B, A, D, end color #ff00af and \orange{\angle{ADB}}∠ADBstart color #ffa500, angle, A, D, B, end color #ffa500 are equal. Since, the angles in a triangle sum to 180^\circ180   180∠ABD+∠BAD+∠ADB∠ABD+2⋅∠ADB∠ABD===180∘180∘180∘−2⋅∠ADB

So, \blue{\angle{ABE}} = 2 \cdot \orange{\angle{ADB}}∠ABE=2⋅∠ADBstart color #6495ed, angle, A, B, E, end color #6495ed, equals, 2, dot, start color #ffa500, angle, A, D, B, end color #ffa500. We can also create an isosceles triangle, \pink{BCD}BCDstart color #ff00af, B, C, D, end color #ff00af.Using the same argument we can find that, \blue{\angle{CBE}} = 2 \cdot \orange{\angle{CDB}}∠CBE=2⋅∠CDBstart color #6495ed, angle, C, B, E, end color #6495ed, equals, 2, dot, start color #ffa500, angle, C, D, B, end color #ffa500.∠ABC=∠ABE−∠CBEstart color #6495ed, angle, A, B, C, end color #6495ed, equals, start color #6495ed, angle, A, B, E, end color #6495ed, minus, start color #6495ed, angle, C, B, E, end color #6495ed

\orange{\angle{ADC}} = \orange{\angle{ADE}} - \orange{\angle{CDE}}∠ADC=∠ADE−∠CDEstart color #ffa500, angle, A, D, C, end color #ffa500, equals, start color #ffa500, angle, A, D, E, end color #ffa500, minus, start color #ffa500, angle, C, D, E, end color #ffa500

Hint #1010 / 11

Since \blue{\angle{ABE}} = 2 \cdot \orange{\angle{ADE}}∠ABE=2⋅∠ADEstart color #6495ed, angle, A, B, E, end color #6495ed, equals, 2, dot, start color #ffa500, angle, A, D, E, end color #ffa500 and \blue{\angle{CBE}} = 2 \cdot \orange{\angle{CDE}}∠CBE=2⋅∠CDEstart color #6495ed, angle, C, B, E, end color #6495ed, equals, 2, dot, start color #ffa500, angle, C, D, E, end color #ffa500, \blue{\angle{ABC}} = 2 \cdot \orange{\angle{ADC}}∠ABC=2⋅∠ADCstart color #6495ed, angle, A, B, C, end color #6495ed, equals, 2, dot, start color #ffa500, angle, A, D, C, end color #ffa500

Hint #1111 / 11

So, \orange{\angle{ADC}} = \frac{1}{2} \cdot \blue{46^\circ} = 23^\circ∠ADC=  

2

1

​  

⋅46  

∘

=23  

∘

start color #ffa500, angle, A, D, C, end color #ffa500, equals, start fraction, 1, divided by, 2, end fraction, dot, start color #6495ed, 46, degrees, end color #6495ed, equals, 23, degrees