# Tetrahedron

• Aug 25th 2009, 05:29 AM
thereddevils
Tetrahedron
In the tetrahedron ABCD , AB = BC =10 , $AC=8\sqrt{2}$ , AD=CD=8 cm and BD = 6 cm . Show that the line from C perpendicular to AB and the line from D perpendicular to AB meet at a point AB .

I can see from the sketch and deduce that they meet at the same point . But how can i show mathematically . THanks !!
• Aug 25th 2009, 07:54 AM
earboth
Quote:

Originally Posted by thereddevils
In the tetrahedron ABCD , AB = BC =10 , $AC=8\sqrt{2}$ , AD=CD=8 cm and BD = 6 cm . Show that the line from C perpendicular to AB and the line from D perpendicular to AB meet at a point AB .

I can see from the sketch and deduce that they meet at the same point . But how can i show mathematically . THanks !!

1. Show that triangle ABD is a right triangle.

2. Calculate DF (=4.8) (it's the height in the right triangle!)

3. Calculate AF ( = 6.4) and BF (=3.6) (use Euclid's theorem)

4. Calculate the length of CF in the right triangle AFC and calculate the length of CF in the right triangle BFC. In both cases you'll get the same result. Therefore the lines DF and CF are perpendicular on AB and meet in the same point.
• Aug 25th 2009, 08:05 AM
HallsofIvy
Quote:

Originally Posted by thereddevils
In the tetrahedron ABCD , AB = BC =10 , $AC=8\sqrt{2}$ , AD=CD=8 cm and BD = 6 cm . Show that the line from C perpendicular to AB and the line from D perpendicular to AB meet at a point AB .

I can see from the sketch and deduce that they meet at the same point . But how can i show mathematically . THanks !!

They are in different planes. If they meet at all, it must be on line AB. Call the point at which the line from D meets AB, P. Then ADP is a right triangle with hypotenuse, AD, of length 8. The lengths of AP and DP are unknown. Call them x and y, respectively. We have $x^2+ y^2= 64$ Similarly BDP is a right triangle with hypotenuse, DB, of length 6. The lengths of BP is unknown. Call it z. Since DP is also a leg in this triangle, we must have $y^2+ z^2= 36$. We must also have x+ z= 10. That gives three equations to solve for x, y, and z.

Do the same with the right triangles APC and BPC and show that x and z, the distances from A to P and from B to P are the same for both problems.