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I need help with these 2 problems:
1)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 $AB/2$.
2)In triangle $ABC$, $AB = AC$, $D$ is the midpoint of $\overline{BC}$, $E$ is the foot of the perpendicular from $D$ to $\overline{AC}$, and $F$ is the midpoint of $\overline{DE}$. Prove that $\overline{AF}$ is perpendicular to $\overline{BE}$.
Can you solve these 2 problems by writing simple geometry proofs? THANK YOU!!
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I have placed the image for #1 in your post but I believe that question has already been posted.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi apsara123,
Please mention the topic on which (2) is based, pure geometry, co ordinate geometry or vectors? It is already solved in co ordinate geometry method in some other thread.
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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See http://www.mathisfunforum.com/viewtopic.php?id=22435&p=2
for problem 1
See the link
http://mathisfunforum.com/viewtopic.php?id=20628
for problem 2
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Hi apsara 123,
You got them O.K.?
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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Pages: 1