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sureshbala problem #1 - triangle sides & angles

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by cbenk121 » Sun Oct 04, 2009 8:45 pm
Had a question on sureshbala's problem #1 in his 780+ thread...

Can we say that the angles opposite two equal chords are equal as well, under any circumstances?

Can we expand this, to say that if you have two triangles sharing a length, and one side in each triangle is equal to the other, that their opposing angles are equal as well?

For example, consider the following triangles sharing length AC (hope it attaches). If AD = BC, then BAC = ACD?

If that's the case, why was the circle needed in the solution?
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by sanjana » Mon Oct 05, 2009 3:28 am
In the Figure attatched,

The 2 triangles share 1 side,and the 2nd side is equal in both traingles,therefore the 3rd side has to be equal.
Hence the 2 triangles are similar and their angles are proportional.(will be equal in this case since the corresponding sides are in the ratio 1:1)

Am i missing something here?
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by cbenk121 » Mon Oct 05, 2009 12:07 pm
sanjana wrote:In the Figure attatched,

The 2 triangles share 1 side,and the 2nd side is equal in both traingles,therefore the 3rd side has to be equal.
Hence the 2 triangles are similar and their angles are proportional.(will be equal in this case since the corresponding sides are in the ratio 1:1)

Am i missing something here?
Why would the third sides have to be equal to each other?

For instance, consider angle CAD = 10 degrees, while angle BCA = 30 degrees. AC = AC, and BC = AD, but BA != CD.

Also...angles in a triangle could not be proportional to each other. Suppose the angles were 1:2. Then the sum of the angles in the larger triangle would add up to 360, an impossibility.
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