ABC is triangle right anled at A.D is perpendicular to BC from A.AB=15,AC=20.P and Q are incentres of triangle

Question

ABC is triangle right anled at A.D is perpendicular to BC from A.AB=15,AC=20.P and Q are incentres of triangles ABD and ADC respectively.Find PQ.

Answer 1

Set up two variables: x = AD, y = BD. Using Pythagoras and a bit of algebra, we find that y = 9 and x = 12. Suppose that the radius of the smaller incircle is 3; does this give consistent results? It does, and since triangles ADC and ABD are in the ratio of 4 to 3, the radius of the larger incircle is 4. Assuming A at the origin, we now have the coordinate points of both incircles as (9, 3) and (4, 8). Finishing the solution is trivial.

Answer 2

E=mcmc

Answer 3

in view that A = 90º, then by Pythagoras Theorem c² + b² = a² a = ?(c² + b²) additionally p² + BD² = c² BD = ?(c² - p²) and p² + DC² = b² DC = ?(b² - p²) yet BD + DC = a, so ?(c² - p²) + ?(b² - p²) = a Equating the above to the 1st equation, we get ?(c² + b²) = ?(c² - p²) + ?(b² - p²) squaring the two facets we get: c² + b² = c² - p² + b² - p² + 2?(c² - p²) (b² - p²) 2p² = 2?(c² - p²) (b² - p²) p² = ?(c² - p²) (b² - p²) p^4 = c²b² - c²p² - b²p² + p^4 c²p² + b²p² = c²b² p²(c² + b²) = c²b² p² = c²b²/(c² + b²) (divide the two facets by a million) a million/p² = (c² + b²)/c²b² a million/p² = c²/c²b²+ b²/c²b² a million/p² = a million/b²+ a million/c² subsequently shown