Suppose $AK$ intersects the circle at point$K,J$,the incircle touches $AC$ at point $L$. Let $AC=b,BC=a,AB=c$. We have $CK=(a+b-c)/2,AL=(b+c-a)/2$. Then from the Pythagorean Theorem we get $AK^2=AC^2+CK^2=b^2+(a+b-c)^2/4$ From $AL^2=AJ*AK$ we get $AJ=AL^2/AK$ While distance from $C$ to $AK$ is $AC*CK/AK$ Just examine $KJ=AK-AJ=2AC*CK/AK$

Consider the foot of the perpendicular from the incenter. Then you get 2 congruent triangles and done.

Let $AK$ meet incircle at $L$ again. We have from similar triangles, \[\frac{CM}{CK}=\frac{AC}{AK}.\]Also suppose $X$ be the $B-$ intouch point. From PoP, $KL=\frac{AK^2-AX^2}{AK}$. Thus we need to show \[2AC\cdot AK=AK^2-AX^2=AC^2+CK^2-AX^2\Leftrightarrow AX^2=(AC-CK)^2=(AC-CX)^2=AX^2.\]

Can someone post a solution using synthetic geometry and complex no.s plsssss

Smita wrote: Can someone post a solution using synthetic geometry and complex no.s plsssss Wizard_32 wrote: Consider the foot of the perpendicular from the incenter. Then you get 2 congruent triangles and done.

TheDarkPrince wrote: Smita wrote: Can someone post a solution using synthetic geometry and complex no.s plsssss Wizard_32 wrote: Consider the foot of the perpendicular from the incenter. Then you get 2 congruent triangles and done. Thanks

Let $AK$ meet the incircle again at $J$ and let the incircle touch $AC$ at $L$. By power of a point, $AL^2=AJ \cdot AK$. Thus $$JK = AK-AJ = AK - \frac{AL^2}{AK} = \frac{AK^2-AL^2}{AK} = \frac{AC^2+CK^2-AL^2}{AK}$$. By similar triangles $\frac{AC}{CH}=\frac{AK}{CK}$ so $2CH=\frac{2AC\cdot CK}{AK}$. It remains to prove $2AC\cdot CK = AC^2+CK^2-AL^2$ or $AL^2 = (AC-CK)^2$ which is true, since $CK=CL$.