# GeometryWatch

#1
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.
0
9 years ago
#2
(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 ..
COnsider the areas of triangle ADC and and BDC
Area of BDC=1/2.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...
0
9 years ago
#3
(Original post by rbnphlp)
the two triangles ADC and BDC are similar ..
Can you justify that statement?
9 years ago
#4
(Original post by ghostwalker)
Can you justify that statement?
Its a sneaky assumption(due to symmetry)
0
9 years ago
#5
(Original post by rbnphlp)
Its a sneaky assumption(due to symmetry)

Intrigued. And what symmetry is that?
9 years ago
#6
(Original post by ghostwalker)
Intrigued. And what symmetry is that?
ok dont crucify me.. I will write up another solution
0
9 years ago
#7
(Original post by rbnphlp)
ok dont crucify me.. I will write up another solution
Lol.
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