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Two circles are externally tangent at point $P$, as shown. Segment $\overline{CPD}$ is parallel to common external tangent $\overline{AB}$. Prove that the distance between the midpoints of $\overline{AB}$ and $\overline{CD}$ is $AB/2$.

I first started with some construction: We start by drawing a common tangent to the two circles passing through $P$ and intersecting $AB$ at $M$. Then we extend $AC$ and $BD$ so that they intersect at $Q$.Next, we extend $QM$ to meet $CD$ at N.

I divided the proof into five steps: 1)$M$ is midpoint of $AB$. 2)$M$ is the circumcenter of $\triangle APB$. 3)$A$ and $B$ are the midpoints of $\overline{CQ}$ and $\overline{DQ}$, respectively. 4) $N$ is the midpoint of $\overline{CD}$. 5)$MN=MP$.

I could prove (1) through (4) and I am not able to prove (5). Can anyone help me out with this?enter image description here

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enter image description here

Let $AE$ and $BF$ be perpendicular to $AB$, with $E$ and $F$ lying on $CD$.

Let $N$ be the reflection of $P$, across the midpoint of $EF$. It follows that $CN=CE+EN=EP+PF=NF+FD=ND$, so $N$ is the midpoint of $CD$.

It follows that $ABFE$ is a rectangle, hence it is similar to $BAEF$.

Applying power of point $M$ to both circles, it can be seen that $MA=MB=MP=\frac{1}{2}AB$.

The reflection bringing $ABFE$ to $BAEF$ maps $M$ to itself and $P$ to $N$.

Hence, $MN=MP=\frac{1}{2}AB$.

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  • $\begingroup$ How can you say $MN=MP$ or how can you say that $P$ maps to $N$ on reflection of $ABFE$ to $BAEF$. Can you please explain me? $\endgroup$ – Gayatri Nov 29 '15 at 8:41
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try this approach to solve the problem. Let me know if you were successful.

enter image description here

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