# geometry, isosceles proofs!

• Feb 18th 2008, 09:07 PM
cbbplanet
geometry, isosceles proofs!

ABCD is a square with AB = 12. Point P is an interior point whose distances from A, B, and CD are equal. Find that distance. Show all work and justify your steps. Re-draw the figure as you explain/justify your solution.

I get stuck after proofing that APB is an isosceles triangle and is similar by ASA postualte
• Feb 18th 2008, 09:19 PM
earboth
Quote:

Originally Posted by cbbplanet

ABCD is a square with AB = 12. Point P is an interior point whose distances from A, B, and are equal. Find that distance. Show all work and justify your steps. Re-draw the figure as you explain/justify your solution.

I get stuck after proofing that APB is an isosceles triangle and is similar by ASA postualte

There doesn't exist an unique solution.

The distances $\overline{AP} = \overline{BP}$ depend from the length of h.

Use Pythagorean theorem:

$d(h) = \sqrt{h^2 + 6^2}~,~ 0 \leq h \leq 12$
• Feb 18th 2008, 09:29 PM
cbbplanet
sorry, i was rereading the question and i guess i left out an important part.

ABCD is a square with AB = 12. Point P is an interior point whose distances from A, B, and CD are equal. Find that distance. Show all work and justify your steps. Re-draw the figure as you explain/justify your solution.

i left out CD and there's an correct answer. the teacher said the answer was 7.5 but we have to show how we solve it
• Feb 18th 2008, 09:40 PM
earboth
Quote:

Originally Posted by cbbplanet
sorry, i was rereading the question and i guess i left out an important part.

ABCD is a square with AB = 12. Point P is an interior point whose distances from A, B, and CD are equal. Find that distance. Show all work and justify your steps. Re-draw the figure as you explain/justify your solution.

i left out CD and there's an correct answer. the teacher said the answer was 7.5 but we have to show how we solve it

You can show that P must be the center of the square. The distance to the vertices of the square is half of a diagonal. So calculate the length of the diagonal first and then divide it by 2:

$D = \sqrt{12^2 + 12^2} = 12 \cdot \sqrt{2}$

And therefore the distance $|\overline{AP}| =|\overline{BP}| =|\overline{CP}| =|\overline{DP}| = 6 \cdot \sqrt{2}$
• Feb 18th 2008, 09:43 PM
cbbplanet
well the problem i'm having is how to prove that point P is in the center of the square.

ps. thanks for being so patient with me.
• Feb 18th 2008, 09:46 PM
earboth
Quote:

Originally Posted by cbbplanet
sorry, i was rereading the question and i guess i left out an important part.

ABCD is a square with AB = 12. Point P is an interior point whose distances from A, B, and CD are equal. Find that distance. Show all work and justify your steps. Re-draw the figure as you explain/justify your solution.

i left out CD and there's an correct answer. the teacher said the answer was 7.5 but we have to show how we solve it

If you mean that the diatnces to A and to B are equal to the (perpendicular) distance to the side (CD) ...

Using my sketch:

distance to (CD): 12 - h
distance (AP): $\sqrt{h^2+6^2}$

Both distances are equal:

$12-h = \sqrt{h^2+6^2}$ . Square both sides of the equation:

$(12-h)^2 = h^2 + 36~\implies~24 h = 108$

I leave the rest for you.
• Feb 18th 2008, 09:52 PM
cbbplanet
thanks so much! i got it =] now i can finally sleep ><