Try this beautiful problem from ISI BStat 2017 Subjective 2 based on right-angled triangle in a circle.
Problem: ISI BStat 2017 Subjective 2
Consider a circle of radius 6 as given in the diagram below. Let B, C, D and E be points on the circle such that BD and CE, when extended, intersect at A. If AD and AE have length 5 and 4 respectively, and DBC is a right angle, then show that the length of BC is $$ \frac {12 + 9 \sqrt {15} }{5} $$
Discussion:
This is a simple application of the power of a point. We have to find the power of point A.
Suppose B = (0,0), C = (x,0) and D = (0,y).
Since AD = 5, therefore, coordinate of A = (0, 5+y).
Using the property of the power of point A, we have: \( AD \times AB = AE \times AC \) . Since AE is 4, this implies
\( 5 \times (5+y) = 4 \times \sqrt { x^2 + (5+y)^2 } \ \Rightarrow 25 (5+y)^2 = 16 (x^2 + (5+y)^2 ) \ \Rightarrow 9(5+y)^2 = 16x^2 \)
Since x and y are positive, we can take square roots on both sides. Hence we will have:
\( 3(5+y) = 4x \ \Rightarrow y = \frac {4x}{3} - 5 \)
Now note that \( \angle DBC = 90^o \) implying DC is a diameter. Hence DC is 12 (as radius is given to be 6).
Hence \( x^2 + y^2 = 12^2 \).
Replacing y by \( \frac {4x}{3} - 5 \) as found earlier, we have
\( \displaystyle {x^2 + \left (\frac {4x}{3} - 5 \right )^2 = 12^2 \ \Rightarrow 9x^2 + 16x^2 - 120x + 225 = 144 \times 9 \ \Rightarrow 25x^2 - 120x - 1071 = 0 \ \Rightarrow x = \frac{120 \pm \sqrt { 120^2 + 107100}}{50} \ \Rightarrow x = \frac {120 \pm \sqrt { 14400 + 107100}}{50} \ \Rightarrow x = \frac {120 \pm 10 \sqrt {1215}}{50} \ \Rightarrow x = \frac {12 \pm 9 \sqrt {15}}{5} } \)
Since x is positive we have \(\displaystyle { x = \frac {12 + 9 \sqrt {15}}{5} } \)
(proved)
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[…] frac {12 + 9 sqrt {15} }{5} $$ ……………………… Discussion […]
There is a more elegant solution available. Using secant theorem sinc=4/5
Hence cosc=3/5.
Also dec is right angled at E so is DEA. In triangle DEA using pythagoras DE=3
In triangle DEC, using pythagoras EC= 3root15.
Hence BC=AC cosc= (4+3root15)*3/5
Proved
sir i have a solution just using classic euclidean geometry.how to send ?
DEC is right angled at E so is DEA. In triangle DEA using Pythagoras DE=3.
NOW ADE & ABC ARE SIMILAR SO , ARE AD/DE=AC/BC. AND AC=5+ROOT(12^2+5^2)=>4+3ROOT15
USING THE RATIO WE GET 4+3ROOT15/BC=5/3 BC=(12+9ROOT15)/5