How to prove that $Delta CFE $ is similar to $Delta CED.$
$begingroup$
Choose a point $C$ outside a circle, and a ray from $C$ cuts the circle at $F$ and $D$. Prove that $Delta CFEsimeqDelta CED.$ $E$ is a point where one of the two tangents from $C$ meets the circle.
Obviously,
$$angle C= angle C$$
I just need to prove that
$$angle CED =angle CFE$$
Or
$$angle CEF =angle CDE$$
I tried to find all the possible angle relationships I could, but I am still not able to prove any of the two conditions above. Could you give me a hint?
geometry
asked Dec 30 '18 at 22:31
LarryLarry
2,55531131
$endgroup$
add a comment |
$begingroup$
Choose a point $C$ outside a circle, and a ray from $C$ cuts the circle at $F$ and $D$. Prove that $Delta CFEsimeqDelta CED.$ $E$ is a point where one of the two tangents from $C$ meets the circle.
Obviously,
$$angle C= angle C$$
I just need to prove that
$$angle CED =angle CFE$$
Or
$$angle CEF =angle CDE$$
I tried to find all the possible angle relationships I could, but I am still not able to prove any of the two conditions above. Could you give me a hint?
geometry
asked Dec 30 '18 at 22:31
LarryLarry
2,55531131
$endgroup$
add a comment |
$begingroup$
Choose a point $C$ outside a circle, and a ray from $C$ cuts the circle at $F$ and $D$. Prove that $Delta CFEsimeqDelta CED.$ $E$ is a point where one of the two tangents from $C$ meets the circle.
Obviously,
$$angle C= angle C$$
I just need to prove that
$$angle CED =angle CFE$$
Or
$$angle CEF =angle CDE$$
I tried to find all the possible angle relationships I could, but I am still not able to prove any of the two conditions above. Could you give me a hint?
geometry
asked Dec 30 '18 at 22:31
LarryLarry
2,55531131
$endgroup$
Choose a point $C$ outside a circle, and a ray from $C$ cuts the circle at $F$ and $D$. Prove that $Delta CFEsimeqDelta CED.$ $E$ is a point where one of the two tangents from $C$ meets the circle.
Obviously,
$$angle C= angle C$$
I just need to prove that
$$angle CED =angle CFE$$
Or
$$angle CEF =angle CDE$$
I tried to find all the possible angle relationships I could, but I am still not able to prove any of the two conditions above. Could you give me a hint?
geometry
geometry
asked Dec 30 '18 at 22:31
LarryLarry
2,55531131
asked Dec 30 '18 at 22:31
LarryLarry
2,55531131
asked Dec 30 '18 at 22:31
LarryLarry
2,55531131
asked Dec 30 '18 at 22:31
LarryLarry
2,55531131
asked Dec 30 '18 at 22:31
LarryLarry
2,55531131
2,55531131
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To show that
$$angle CEF =angle CDE$$
draw the diameter $EE'$ which is clearly perpendicular to $CE$ and hence
$$angle CEF +angle FEE'=90^circ$$
and hence
$$
angle EE'F =angle CEF
$$
But
$$
angle EE'F=angle CDE
$$
since they are inscribed angles looking at the same arc.
Hence
$$angle CEF =angle CDE$$
answered Dec 30 '18 at 22:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.8k1385165
$endgroup$
add a comment |
$begingroup$
Note:
$$CE^2=CFcdot CD iff frac{CF}{CD}=left(frac{CE}{CD}right)^2.$$
and:
$$S_{Delta CEF}=frac12cdot CEcdot CFcdot sin angle ECF; \
S_{Delta CDE}=frac12cdot CEcdot CDcdot sin angle ECD; \
frac{S_{Delta CEF}}{S_{Delta CDE}}=frac{CF}{CD}=left(frac{CE}{CD}right)^2 Rightarrow Delta CEF simDelta CDE.$$
answered Dec 30 '18 at 22:51
farruhotafarruhota
22.5k2942
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
To show that
$$angle CEF =angle CDE$$
draw the diameter $EE'$ which is clearly perpendicular to $CE$ and hence
$$angle CEF +angle FEE'=90^circ$$
and hence
$$
angle EE'F =angle CEF
$$
But
$$
angle EE'F=angle CDE
$$
since they are inscribed angles looking at the same arc.
Hence
$$angle CEF =angle CDE$$
answered Dec 30 '18 at 22:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.8k1385165
$endgroup$
add a comment |
$begingroup$
To show that
$$angle CEF =angle CDE$$
draw the diameter $EE'$ which is clearly perpendicular to $CE$ and hence
$$angle CEF +angle FEE'=90^circ$$
and hence
$$
angle EE'F =angle CEF
$$
But
$$
angle EE'F=angle CDE
$$
since they are inscribed angles looking at the same arc.
Hence
$$angle CEF =angle CDE$$
answered Dec 30 '18 at 22:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.8k1385165
$endgroup$
add a comment |
$begingroup$
To show that
$$angle CEF =angle CDE$$
draw the diameter $EE'$ which is clearly perpendicular to $CE$ and hence
$$angle CEF +angle FEE'=90^circ$$
and hence
$$
angle EE'F =angle CEF
$$
But
$$
angle EE'F=angle CDE
$$
since they are inscribed angles looking at the same arc.
Hence
$$angle CEF =angle CDE$$
answered Dec 30 '18 at 22:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.8k1385165
$endgroup$
To show that
$$angle CEF =angle CDE$$
draw the diameter $EE'$ which is clearly perpendicular to $CE$ and hence
$$angle CEF +angle FEE'=90^circ$$
and hence
$$
angle EE'F =angle CEF
$$
But
$$
angle EE'F=angle CDE
$$
since they are inscribed angles looking at the same arc.
Hence
$$angle CEF =angle CDE$$
answered Dec 30 '18 at 22:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.8k1385165
answered Dec 30 '18 at 22:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.8k1385165
answered Dec 30 '18 at 22:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.8k1385165
answered Dec 30 '18 at 22:54
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.8k1385165
63.8k1385165
add a comment |
add a comment |
$begingroup$
Note:
$$CE^2=CFcdot CD iff frac{CF}{CD}=left(frac{CE}{CD}right)^2.$$
and:
$$S_{Delta CEF}=frac12cdot CEcdot CFcdot sin angle ECF; \
S_{Delta CDE}=frac12cdot CEcdot CDcdot sin angle ECD; \
frac{S_{Delta CEF}}{S_{Delta CDE}}=frac{CF}{CD}=left(frac{CE}{CD}right)^2 Rightarrow Delta CEF simDelta CDE.$$
answered Dec 30 '18 at 22:51
farruhotafarruhota
22.5k2942
$endgroup$
add a comment |
$begingroup$
Note:
$$CE^2=CFcdot CD iff frac{CF}{CD}=left(frac{CE}{CD}right)^2.$$
and:
$$S_{Delta CEF}=frac12cdot CEcdot CFcdot sin angle ECF; \
S_{Delta CDE}=frac12cdot CEcdot CDcdot sin angle ECD; \
frac{S_{Delta CEF}}{S_{Delta CDE}}=frac{CF}{CD}=left(frac{CE}{CD}right)^2 Rightarrow Delta CEF simDelta CDE.$$
answered Dec 30 '18 at 22:51
farruhotafarruhota
22.5k2942
$endgroup$
add a comment |
$begingroup$
Note:
$$CE^2=CFcdot CD iff frac{CF}{CD}=left(frac{CE}{CD}right)^2.$$
and:
$$S_{Delta CEF}=frac12cdot CEcdot CFcdot sin angle ECF; \
S_{Delta CDE}=frac12cdot CEcdot CDcdot sin angle ECD; \
frac{S_{Delta CEF}}{S_{Delta CDE}}=frac{CF}{CD}=left(frac{CE}{CD}right)^2 Rightarrow Delta CEF simDelta CDE.$$
answered Dec 30 '18 at 22:51
farruhotafarruhota
22.5k2942
$endgroup$
Note:
$$CE^2=CFcdot CD iff frac{CF}{CD}=left(frac{CE}{CD}right)^2.$$
and:
$$S_{Delta CEF}=frac12cdot CEcdot CFcdot sin angle ECF; \
S_{Delta CDE}=frac12cdot CEcdot CDcdot sin angle ECD; \
frac{S_{Delta CEF}}{S_{Delta CDE}}=frac{CF}{CD}=left(frac{CE}{CD}right)^2 Rightarrow Delta CEF simDelta CDE.$$
answered Dec 30 '18 at 22:51
farruhotafarruhota
22.5k2942
edited Dec 31 '18 at 0:06
answered Dec 30 '18 at 22:51
farruhotafarruhota
22.5k2942
answered Dec 30 '18 at 22:51
farruhotafarruhota
22.5k2942
answered Dec 30 '18 at 22:51
farruhotafarruhota
22.5k2942
22.5k2942
add a comment |
add a comment |
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