SOLUTION: This is on Ptolemy's theorem. If CD =1, BC = 2, AD = 3, AB = 4, and the triangle ADC is a right triangle, can the quadrilateral ABCD be inscribed in a circle? Yes or No

Algebra ->  Triangles -> SOLUTION: This is on Ptolemy's theorem. If CD =1, BC = 2, AD = 3, AB = 4, and the triangle ADC is a right triangle, can the quadrilateral ABCD be inscribed in a circle? Yes or No      Log On


   



Question 33829: This is on Ptolemy's theorem. If CD =1, BC = 2, AD = 3, AB = 4, and the triangle ADC is a right triangle, can the quadrilateral ABCD be inscribed in a circle?
Yes or No

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
This is on Ptolemy's theorem. If CD =1, BC = 2, AD = 3, AB = 4, and the triangle ADC is a right triangle, can the quadrilateral ABCD be inscribed in a circle?
TO USE PTOLEMYS THEOREM YOU NEED AC AND BD AND CHECK IF
AB*CD+AD*BC=AC*BD...IF YES IT IS CYCLIC QUADRILATERAL OTHERWISE NOT..HERE WE CAN FIND AC..BUT BD NEEDS CALCULATION..SO I AM GIVING ANOTHER WAY TO DO IT FROM PROPERTY OF CYCLIC QUADRILATERAL THAT SUM OF OPPOSITE ANGLES=180
ADC IS RIGHT TRIANGLE...ANGLE D=90...SO....AC^2=AD^2+DC^2=3^2+1^2=10
WE FIND THAT IN TRIANGLE ABC
AB^2+BC^2=4^2+2^2=20 WHICH IS NOT EQUAL TO AC^2...HENCE ABC IS NOT A RIGHT TRIANGLE.HENCE ANGLE B IS NOT 90.
HENCE SUM OF OPPOSITE ANGLES ADC AND ABC IS NOT 180..HENCE ABCD IS NOT A CYCLIC QUADRILATERAL...