pls help me on this;
two circles touch externally at T. A chord of the first circle XY is produced and touches the other at Z. The chord ZT of the second circle. when produced, cuts the first circle at W. prove that angle XTW IS equal to angle YTZ
pls help me on this;
two circles touch externally at T. A chord of the first circle XY is produced and touches the other at Z. The chord ZT of the second circle. when produced, cuts the first circle at W. prove that angle XTW IS equal to angle YTZ
Hello josemalawiDraw the common tangent at T, meeting YZ at A.
Let angle ATZ = $\displaystyle \theta$, and angle ATY = $\displaystyle \phi$
$\displaystyle \therefore$ angle YTX = $\displaystyle \theta + \phi$
Now AT = AZ (tangents from A to the circle)
$\displaystyle \therefore$ angle AZT = $\displaystyle \theta$ (angles in isosceles triangle)
$\displaystyle \therefore$ angle XYT = $\displaystyle 2\theta + \phi$ (exterior angle of triangle YTZ)
Angle YXT = angle YTA = $\displaystyle \phi$ (angle in alternate segment)
$\displaystyle \therefore$ angle XTY = $\displaystyle 180 - (2\theta + 2\phi)$ (angle sum of triangle XYT)
$\displaystyle \therefore$ angle XTW = 180 - angle XTY - angle YTZ (angles on a straight line)
$\displaystyle = 180 - (180 -(2\theta + 2\phi)) - (\theta + \phi)$
$\displaystyle = \theta + \phi$
= angle YTZ
QED
Grandad