11/01/2025

The diameter $AC$ and $BD$ of two semi-circles is tangent to the other semi-circle at mean time. Prove that $AB \parallel CD$.

Prove:

\(\begin{multline} \shoveleft \text{Let }O_1, O_2 \text{ be the center of semi-circle }BD \text{ and }AC \text{ respectively}\\ \shoveleft \text{And }E, F\text{ are the tangent point on diameter }BD, AC \text{ respectively} \implies \\ \shoveleft O_1F\perp AC, O_2E\perp BD\implies O_1EFO_2\text{ are cyclic} \implies \angle{BO_1F}=\angle{AO_2E}\\ \shoveleft BO_1=FO_1, AO_2=EO_2 \implies \angle{O_1BF}=90^{\circ}-\dfrac{BO_1F}{2}=90^{\circ}-\dfrac{AO_2E}{2}=\angle{EAO_2}\\ \shoveleft \implies EBAF\text{ are cyclic} \implies \angle{AFB}=\angle{AEB}\implies \angle{ACE}=\angle{BDF}\\ \shoveleft \implies DEFC \text{ are cyclic} \implies \angle{FCD}=\angle{BEF}=180^{\circ}-\angle{BAF}\implies AB\parallel CD \blacksquare \end{multline}\)

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