the bisector of angle a meets bc at u if ux is drawn parallel to ac meeting ab at x and uy drawn parallel to ab meets bc at e prove that bx/cy = absquare/acsquare
Answers
Answer: Yes it can be proved easily just you need to remember some points from trigonometry and similarity of triangles .
Step-by-step explanation:
Consider the given image along attachment,
There we can assume the angles as
ang.XBU=a
ang.XUB=b
ang.YUC=c
ang.YCU=d
Now as the sides UY//AB and XU//AC with ang.A bisected so by properties of parallel lines , in quad.AXUY,
ang.(A=X=U=Y) ,hence all the four sides become equal making the given quadrilateral as square so,
AX=XU=UY=AY
Now in ∆XUB and ∆YUC,
we can say that, b =(90-c) , and , c=(90-b)
so,. sin b=sin(90-c) and, sin c=sin(90-b)
BX/BU =cos c and, CY/CU = cos b
BX/BU = UY/UC and, CY/CU= UX/BU
Divide both the equations ,
BX/BU =UY/UC
CY/CU =UX/BU
from above equation, BX = BU² (1)
CY CU²
By using the property of angule bisector we get to know that if it divides any angle then it also makes a ratio as ,
BU/CU = AB/AC
Squaring both the sides we get,
BU²/CU² = AB²/AC² (2)
Using equation 1 and 2 we get ,
BX = AB²
CY AC²
proved
