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sh00ter
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#1
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Triangle ABC (angle ACB > 90) is inscribed in a circle centered at S. D is the intersection point of segments CS and AB. Prove that if AC+BC=2CS, then radii of the circles inscribed into triangles ADC and BDC are equal.
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rbnphlp
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(Original post by sh00ter)
Triangle ABC (angle ACB > 90) is inscribed in a circle centered at S. D is the intersection point of segments CS and AB. Prove that if AC+BC=2CS, then radii of the circles inscribed into triangles ADC and BDC are equal.
the two triangles ADC and BDC are similar ..
so AD/BD=DC/DC=AC/BC
so AD/BD=1---(a)
COnsider the areas of triangle ADC and and BDC
Area of ADC=1/2.DC.AD
Area of BDC=1/2.DC.BD
Area of ADC/Area of BDC=\frac{0.5.DC.AD}{0.5.DC.BD}
We know triangle AD/BD=1 from (a)

so Area of ADC/Area of BDC=1
therefore Area of ADC=Area of BDC
Since the triangles are similar and have the same area the circle inscribed in
it must have the same radius...
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ghostwalker
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(Original post by rbnphlp)
the two triangles ADC and BDC are similar ..
Can you justify that statement?
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rbnphlp
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(Original post by ghostwalker)
Can you justify that statement?
Its a sneaky assumption(due to symmetry)
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ghostwalker
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(Original post by rbnphlp)
Its a sneaky assumption(due to symmetry)

Intrigued. And what symmetry is that?
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rbnphlp
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(Original post by ghostwalker)
Intrigued. And what symmetry is that?
ok dont crucify me.. I will write up another solution
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ghostwalker
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(Original post by rbnphlp)
ok dont crucify me.. I will write up another solution
Lol.
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