(An interesting problem inspired by this one but still different. And, no, I'm not looking for your help to solve a detail here in order to provide a full solution elsewhere. I'll stop here).
A right-angled triangle ABM is given ($\angle M = 90^\circ)$. A tangent $t$ is drawn at point $M$ to the circumscribed circle with center $O$. Perpendicular bisector $OE$ of side $AM$ meets tangent $t$ at point $D$. Prove that line $DB$ cuts triangle height $MH$ in half ($CM=CH$)
Ok, one way to solve this is to use analytical geometry. It's doable but tedious. More ellegant way is to solve the problem by using complex geometry and I will demonstrate that solution here:
We'll take circumscribed circle to be unit circle. Let's assign the following complex numbers to various points: $z_M=z,z_0=0,z_A=-1,z_B=1$. It's a well know formula that:
$$z_D=\frac{2z_Mz_A}{z_M+z_A}=\frac{-2z}{z-1}$$
Also notice that:
$$Re(C)=Re(M)=\frac{z+\bar{z}}{2}$$
$$Im(C)=\frac12 Im(M)=\frac12\frac1{2i}(z-\bar{z})$$
$$z_C=Re(C)+i\cdot Im(C)=\frac{z+\bar{z}}{2}+\frac14(z-\bar{z})=\frac 14(3z+\bar z)$$
The last thing we have to prove is that points $D,C,B$ are colinear which is true if:
$$z_D-z_B=k(z_C-z_B)$$
…for some real $k$. In other words:
$$\frac{-2z}{z-1}-1=k(\frac 14(3z+\bar z)-1)\iff$$
$$\frac{-2z-(z-1)}{z-1}=\frac{k}{4}(3z+\bar z-4)\iff$$
$$\frac{-3z+1}{z-1}\frac{\bar z-1}{\bar z-1}=\frac{k}{4}(3z+\bar z-4)\iff$$
$$\frac{-3z+1}{z-1}\frac{\bar z-1}{\bar z-1}=\frac{k}{4}(3z+\bar z-4)\iff$$
$$\frac{3z+\bar z-4}{l}=\frac{k}{4}(3z+\bar z-4)\iff$$
$$k=\frac4l, \quad l=(z-1)(\bar z – 1)\in R$$
Question: I think we can prove this with Euclid too, without complex or any other numbers. Can someone come up with one such solution?
Best Answer
extend $BM$ until it meets the line $AD$ at $X$ since $\angle AMX=90$
$AD$ is $\frac{AX} {2}$ by homothety $CH$ is $\frac{MH} {2}$