Suppose, the inradius of \triangle{ABC} is r and the circumradius of \triangle{ABC} is R.
If the circumcenter of \triangle{ABC} is located on AB, then ABC is right-angled triangle, AB is the hypotenuse of \triangle{ABC} and the circumcenter of \triangle {ABC} is on the midpoint of AB
.Given that, AB=29 , AC = 20.
According to Pythagoras theorem, AC^2+BC^2=AB^2
=> 20^2 + BC^2 = 29^2
=> BC^2 = 841-400
=> BC^2 = 441
=> BC = 21
Now, AB=29 , BC=21,AC=20,
\therefore The area of \triangle{ABC}=\cfrac{1}{2}*BC*AC=\cfrac{1}{2}*21*20=210
Again, s =\cfrac{AB+BC+AC}{2}= \cfrac{29+21+20}{2}=35
The length of inradius, r = \cfrac{area\space{of}\space{\triangle{ABC}}}{s} =\cfrac {210}{35}=6
The length of circumradius, R=\cfrac{29}{2}=14.5
The distance of incenter and circumcenter = \sqrt{R^2-2rR} = \sqrt{(14.5)^2-2*6*14.5} = \sqrt{210.25-174}=\sqrt{36.25}=6.020797.......
The distance of incenter and circumcenter = 6.02\fbox{Ans}
Let M be the center of the circle passing through B, P, O, and C.
Because P is the orthocenter of ABC, the two segments AB and BP are hypotenuses for two right triangles in the figure above. Using those triangles, we find that \angle BAC and \angle BPD, shown above in dark green, are both complementary to \angle ABP and therefore congruent to each other.
The Inscribed Angle Theorem does the heavy lifting:
(1) \triangle BPC \cong \triangle BOC because both are inscribed in a circle about M.
(2)\angle BAC has half their measure, as it’s the inscribed angle corresponding to central angle \angle BOC.
Now \angle BPD is supplementary to \angle BPD and has half its measure. Therefore \angle BPD has measure 60^{\circ} , as must \angle BAC .
এখানে, প্রতিটি ধাপে ষড়ভুজ সমান অনুপাতে ছোট হচ্ছে এবং একটা সময় তা শুন্য হয়ে যাবে। কিন্তু, কখনোই ষড়ভুজের আকার এর বাহুর উপর আঁকা বর্গের আকারের চেয়ে ছোট হয় না।
অতএব, এমন কোনো ষড়ভুজ নেই, যার প্রতিটি শীর্ষবিন্দুই বর্গাকার গ্রিডের বিন্দু। [প্রমাণিত]
Let n \geq 2 be a positive integer, and let A, B, and C be points in n-dimensional space. Given that \angle ABC=\angle ACB, prove or disprove that AB=AC. (Problem 17)
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