I have yet another question on triangles? I will not tolerate'Do your homework'as an answer. Read on ...

Question

If two isosceles triangles have a common base, then prove that the line joining their vertices bisects them at right angles.

Answer 1

Let the trianglas be ABC & BCD
AD forms 2 triangles, ABD & ACD
AB=AC since ABC is isisceles
BC+BD since BCD is isosceles.
angle ABC=angle ACB
angle DBC=angls DCB so angls ABD=angle ACD so ABD is congruent to ACD by SAS

call the point where AD crosses BC E

so both vertex angles are bisected.
look at ABC.
AE bisects BAC & AE is a commin side for BAE & CAE.
since AB=AC, the triangles are congruent by SAS
therefore BE=CE
& angle AEB= angle AEC which are supplements so they must be right angles.
QED

Answer 2

If I understand your question correctly 2 isosceles triangles side by side have the same base and an imaginary verticle line which joins with the two vertices. This line would from a 90 degree angle between both bases thus creating two right triangles with the common base. The lengths of each side can be found with a^2 + b^2 = c^2, with a = the common base and b = the height of one of the original triangles if not common among them. The angles must all add up to 180 with angle b = 90 and angle a = 1/2 of the intersecting angle of the isoceles triangle opposite angle b of the right triangle.

So with a common base of 4 and a height of 3 the unknown side or hypotenuse of this newly created right triangle equals 5. And based on the assumption that all angles of the isoceles triangles are 60 degrees therefore 1/2 of 60 = 30 and 90 + 30 = 120 so the unknown angle is 60.

Answer 3

Do your homework, but do it the other way around... Prove that the median of each triangle 1) bisects the base at a right angle, and thus 2) is on the same line joining the vertices.

Answer 4

Drawing the line between their vertices and continuing the line
to the base constructs an upper pair of triangles with equal coresponding sides. So they are congruent and therefore have
equal corresponding angles, 2 of which are at the original vertex
of one of the isosceles triangles. The proof by congruency can be
made for the lower pair of triangles. Congruency by SSS.

Answer 5

I’d suggest something below. Assume ABC is isosceles triangle with CA=CB & AB its base; assume ABC’ is isosceles triangle with C’A=C’B & AB its base, point C’ being on opposite side of AB; assume C’A=CA (=CB=C’B). Now we have a rhombus ACBC’. Its diagonals AB & CC’ are normal to each other, intersecting at point D, i.e. angle CDB=angle C’DB=90degrees. Now take any point C’’ belonging to CC’. A new triangle ABC’’ will be isosceles. If C’’ does not belong to CC’ the triangle ABC’’ will not be isosceles! I guess it proves.

Answer 6

Hmmm... it is POSSIBLE that this line joining their vertices does NOT bisect them and therefore as stated.. it cannot be proven...


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I can't draw it here, but imagine that both of the triangles have the same base (not one base smaller than the other), the line joining the vertices does not bisect the bases or the triangles. And ONE solution that does not bisect makes a proof impossible.

Answer 7

Hi, if you check the link below you will see graphics and it is much easier to explain.


Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD.
I say that DB is to DC as AB is to AC.

Draw CE through C parallel to DA, and carry AB through to meet it at E. I.31
Then, since the straight line AC falls upon the parallels AD and EC, the angle ACE equals the angle CAD. I.29
But the angle CAD equals the angle BAD by hypothesis, therefore the angle BAD also equals the angle ACE.
Again, since the straight line BAE falls upon the parallels AD and EC, the exterior angle BAD equals the interior angle AEC. I.29
But the angle ACE was also proved equal to the angle BAD, therefore the angle ACE also equals the angle AEC, so that the side AE also equals the side AC. I.6
And, since AD is parallel to EC, one of the sides of the triangle BCE, therefore, proportionally DB is to DC as AB is to AE. VI.2
But AE equals AC, therefore DB is to DC as AB is to AC. V.7

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Next, let DB be to DC as AB is to AC. Join AD.

I say that the straight line AD bisects the angle BAC.

With the same construction, since DB is to DC as AB is to AC, and also DB is to DC as AB is to AE, for AD is parallel to EC, one of the sides of the triangle BCE, therefore also AB is to AC as AB is to AE. VI.2
V.11

Therefore AC equals AE, so that the angle AEC also equals the angle ACE. V.9
I.5
But the angle AEC equals the exterior angle BAD, and the angle ACE equals the alternate angle CAD, therefore the angle BAD also equals the angle CAD. I.29
Therefore the straight line AD bisects the angle BAC.
Therefore, if an angle of a triangle is bisected by a straight line cutting the base, then the segments of the base have the same ratio as the remaining sides of the triangle; and, if segments of the base have the same ratio as the remaining sides of the triangle, then the straight line joining the vertex to the point of section bisects the angle of the triangle.

Answer 8

Ok, don't do your homework. Buy yourself a couple of beers, get real drunk, and fail your test.

Answer 9

Kind of demanding aren't we?

Answer 10

Well, if you are going to be a prick about it..forget it