<s>[INST] geometry question - Prove that $KB = KD$ Let $\Delta ABC$ have $AB = AC$. On the line perpendicular to $AC$ at $C$ take $D$ so that points $B, D$ is different side $AC$. Let $K$ is the intersection of the straight line through $B$ perpendicular to $AB$ and the line through the center of $M$ of the $CD$ perpendicular to $AD$. Prove that $KB = KD$

 [/INST] Let $AB=AC=a$, $CD=2b$ and $ML\perp AD$, where $L\in AD$.
Hence, $MD=b$ and since $\Delta MLD\sim \Delta ACD$, we obtain:
$$\frac{LD}{CD}=\frac{MD}{AD}$$ or
$$\frac{LD}{2b}=\frac{b}{\sqrt{a^2+4b^2}},$$ which gives
$$LD=\frac{2b^2}{\sqrt{a^2+4b^2}}.$$
Hence, $$AL=AD-LD=\sqrt{a^2+4b^2}-\frac{2b^2}{\sqrt{a^2+4b^2}}=\frac{a^2+2b^2}{\sqrt{a^2+4b^2}}.$$
Id est, $$KD^2=KL^2+LD^2=AK^2-AL^2+LD^2=BK^2+AB^2-AL^2+LD^2=$$
$$=BK^2+a^2-\left(\frac{a^2+2b^2}{\sqrt{a^2+4b^2}}\right)^2+\left(\frac{2b^2}{\sqrt{a^2+4b^2}}\right)^2=BK^2$$
and we are done!
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