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1) Let's say angle ADC is 30 making angle ACB = 60. That would make triangle ABC a 30, 60, 90 triangle so the answer would be NO. However, ADC can be 22.5 or any angle for that matter but let's say it's 22.5 which makes angle ACB = 45. That would make triangle ABC an isosceles right triangle with the answer YES. This is NOT sufficient.

2) If angle ACB is twice as large as angle CAB we can write it as ACB = 2CAB. We know that angle ABC = 90 degrees so we use the following equations.

ACB = 2CAB

ABC + ACB + CAB = 180 (let's put 90 for ABC since we know that angle) 90 + ACB + CAB = 180 (we can plug in 2CAB in place of ACB from here to get CAB's angle)

90 + 2CAB + CAB = 180 90 + 3CAB = 180 Solving for CAB we get 30 degrees

Triangle ABC can ONLY be a 30, 60, 90 triangle and the answer will always be NO so selection (2) is sufficient.
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I'm trying to not just answer the problem but to explain how I came up with my answer. If I am incorrect or you have a better method please PM me your thoughts. Thanks!

B. s1 - if outside angle is 27.5 than inside angle is 45 and hence triangle is isoceles. However for any other value of outside angle this is not possible. S2 - clearly states triangle cannot be isoceles as the two angles néed to be equal to each other in aright angled triangle

I have a doubt . from crejoc's reply I get that angle CAD= angle ADC . Now considering the triangle ADC the exterior angle ACB = sum of angles( CAD+ CDA) . So angle ACB > angle ADC . Now taking the rule angle opposite to greater side is greater and comparing the angles ACB and ADC we have AB> AC . Hence ABC is not isosceles.

So should n't the answer be D instead of B.... please help...

I have a doubt . from crejoc's reply I get that angle CAD= angle ADC . Now considering the triangle ADC the exterior angle ACB = sum of angles( CAD+ CDA) . So angle ACB > angle ADC . Now taking the rule angle opposite to greater side is greater and comparing the angles ACB and ADC we have AB> AC . Hence ABC is not isosceles.

So should n't the answer be D instead of B.... please help...

AB is a leg in a right triangle ABC and it cannot be more than the hypotenuse AC of the same right triangle.

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In a triangle ABD, B is the right angle. Point C is on BD. If you draw a line from A to C, then is triangle ABC an isosceles triangle?

Notice that since ABC is a right triangle then the question basically asks whether angle ACB is equal to angle CAB, because in this case AB would be equal to BC, thus triangle ABC would be an isosceles triangle.

(1) Angle ACB is twice as large as angle ADC --> it certainly possible angle ACB to be equal to angle CAB (45 degrees) and angle ADC to be 22.5 degrees but it's also possible angle ACB NOT to be equal to angle CAB, for example if angle ACB is 60 degrees and angle ADC is 30 degrees. Not sufficient.

(2) Angle ACB is twice as large as angle CAB --> directly says that angle ACB is NOT equal to angle CAB, hence triangle ABC is NOT isosceles. Sufficient.

to prove triangle ABC is isosceles, we need to prove ACB =BCA=45, This can't be found using option A (on the contrary I came to d conclusion that triangle ACD is isosceles :D)

from option B it's obvious that in triangle ABC, angle ACB is not equal to BAC so triangle ABC can't be isosceles