MHB How can the perpendicular diagonals of a triangle help prove HCD?

[slwarrior64]

 

[slwarrior64]

I have so far that triangles BDE and FDC are similar, as are DCE and DBF

 

[Opalg]

The little triangle $DEG$ is also similar to $BDE$ and $FDC$. Use that to deduce that both of the triangles $DHC$ and $DHF$ are isosceles. The result should then follow quite easily.

 

[slwarrior64]

The little triangle $DEG$ is also similar to $BDE$ and $FDC$. Use that to deduce that both of the triangles $DHC$ and $DHF$ are isosceles. The result should then follow quite easily.

One quick question, is DEG similar to BDE because we know both have a right angle, and then they share a side and they share an angle? Is that enough?

 

[skeeter]

two equal angles are enough to show similarity

 

[slwarrior64]

two equal angles are enough to show similarity

True, I was thinking congruence. Thanks everyone!

 

[slwarrior64]

The little triangle $DEG$ is also similar to $BDE$ and $FDC$. Use that to deduce that both of the triangles $DHC$ and $DHF$ are isosceles. The result should then follow quite easily.

Thank you!

 

[slwarrior64]

The little triangle $DEG$ is also similar to $BDE$ and $FDC$. Use that to deduce that both of the triangles $DHC$ and $DHF$ are isosceles. The result should then follow quite easily.

I'm sorry, I'm definitely missing something similar, but how do I know they are isosceles?

 

[Opalg]

I'm sorry, I'm definitely missing something similar, but how do I know they are isosceles?

The angles $GDE$ and $DFH$ are equal (from the similar triangles $GDE$ and $DFC$).
The angles $GDE$ and $FDH$ are equal (from the intersection of the lines $EF$ and $GH$).
Therefore the angles $DFH$ and $FDH$ are equal and so the triangle $FDH$ is isosceles. So the sides $HF$ and $HD$ must be equal.
Using the fact that the triangle $CDF$ has a right angle at $D$, you can then show that the triangle $HCD$ has equal angles at $C$ and $D$ and is therefore also isosceles.

 

[slwarrior64]

The angles $GDE$ and $DFH$ are equal (from the similar triangles $GDE$ and $DFC$).
The angles $GDE$ and $FDH$ are equal (from the intersection of the lines $EF$ and $GH$).
Therefore the angles $DFH$ and $FDH$ are equal and so the triangle $FDH$ is isosceles. So the sides $HF$ and $HD$ must be equal.
Using the fact that the triangle $CDF$ has a right angle at $D$, you can then show that the triangle $HCD$ has equal angles at $C$ and $D$ and is therefore also isosceles.

Thanks, I was able to get triangle FDH, but I was stuck on proving HCD because I didn't know how that line would cut the right angle

 

[slwarrior64]

Thanks, I was able to get triangle FDH, but I was stuck on proving HCD because I didn't know how that line would cut the right angle

I got it! Thanks again!