Good question...
Given two triangles ABC and DEF, if they have one internal angle from each triangle equal, the opposite sides of these(angles)
equal ,and the corresponding height relative to the mentioned side equal. Then they are congruent.
How do we prove it.
I would like to discuss this interesting property and the different methods to tackle this problem.
I've found out two different methods-both of them related to circumferences-that will be posted as soon
as I gather and order the information.
Please let me know if I'm in the wrong topic, it is not my intention to irritate anyone.
Any comment will be taken into account.
given:
The final intention is to prove:
Theorem:
In any convex quadrilateral ABCD. If any angle( ). made by any of the two diagonal and any side(b), is equal to another angle which is made by the remaining diagonal and the opposite side to b.
then, the it is a tangential quadrilateral.
in order to prove this theorem the triangles
have to be analyzed:
It is true that:
They have two identical angles, which implies that their angles are all identical
If we analyze the triangles:
Two sides have lengths in the same ratio, and the angles included between these sides have the same measure
In the quadrilateral, opposite angles add to 180 degrees, thus it is a tangential quadrilateral.
Back to our problem; if the triangles share one more angle measure ,then , they would be similar. But because they have one corresponding side in common, they would be indeed congruent. ...(*)
The next figure shows the triangles with their base sides one over the other, so the points A and D occupy the same position, likewise C and F
If we draw the line BE:
Becouse they are Alternate Interior Angles.
Also, The quadrilateral ABEC is tangential......(see theorem above)
from (III) and (IV), we can conclude that:
. .....(*)
I will try to post, as soon as possible, a second method(rather intuitive) to prove this problem.
Any comment will be appreciated.
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