Question: Triangles with Shared Vertex

Comment on Triangles with Shared Vertex

straight away we can conclude the y must be greater since it forms line with length 5. x will be smaller compare to y, since it forms line of length 3.
greenlight-admin's picture

I'm not quite sure what you mean by "x will be smaller compare to y, since it forms line of length 3"

consider line A with length 3 units, line B with length 5 units.

If we have projection of endpoints as a line to form an angle ( analogy: in case
of circle, we have inscribed angle for a chord ), when line is smaller, obviously
angle formed will be lesser, compare to the angle formed by larger line.

therefore angle x is smaller than angle y, since angle x is formed by line 3 units and angle y is formed by line 5 units.
greenlight-admin's picture

Sorry, It's still not 100% clear to me. Are you referring to the rule stated at 0:48?


It would be better if i can upload the diagrams.

My assumption was, in general if we have 2 lines with length difference. say one is 2 and other is 5. will project lines at both endpoints with same angle. Those projected lines will meet at some point, forms arbitrary angle. on that case we can tell smaller length line would form smaller angle and larger one would form larger angle.

correct me, if anything wrong.
greenlight-admin's picture

I think that sounds correct :-)

Is this assumption correct?: Without noticing that those two lines are parallel, I noticed that angle x corresponds with side length 2, and that angle y corresponds with side length 5, because 2<5, x must be less than y.
greenlight-admin's picture

The rule you are applying applies to ONE triangle, but you are using it for TWO triangles. In a single triangle, the angle opposite the longest side will be the greatest angle.

However, we can't then apply this concept to compare an angle in one triangle with an angle in a different triangle.

angle x = angle X. Vertically opposite angles. Now 2 angles in both triangles are equal. So the third angle will also be equal. Since 3 angles are equal in both the triangles, I used similar trioangles concept and used the same rule as yours later (longest side is opposite to longest angle and solved.) got the same result. Is this approach correct?
greenlight-admin's picture

Yes, that approach is correct.

Also why haven't you compared the sides and angles of other triangle? You have just compared one triangle (for which all sides and all angles are known).
greenlight-admin's picture

Once we know that triangle ABC has sides with length 2, 3 and 4, we can just use the triangle property described 1:02 at in the following video:

Not this, I meant you have used the triangle property only for triangle ABC and not for triangle EDC, why so?
greenlight-admin's picture

Which triangle property are you referring to?
At 0:32 in the video, we conclude that, since ∠ABC = ∠CED, line AB is PARALLEL to line DE

Once we know that AB || DE, we can see that ∠BAC is equal to ∠CDE (at 0:37 in video)

At this point, we have labeled all 3 angles in ∆ABC.
At this point, we COULD use similar triangle to find the lengths of the sides on ∆CDE, but there's really no need to, since ∆ABC already provides all of the information needed to answer the question.

Hi Brent How will this approach work-

I see BE and AD are intersecting at C so from RULE-
"Opposite Angles are equal for intersection lines"

I say <BCA = <DCE =x

Now both Triangles have both the angles in common x,k So I say they are similar. From the RULE, "When 2 angles in similar triangles are equal then their 3rd angle must also be equal".

I conclude that left over 3rd angle in Triangle CBA is y.

Now In Triangle CBA I have all 3 angles and side So from RULE, "The greatest side in a triangle will have greatest opposite Angle and vice versa". Finally y is grater than x

greenlight-admin's picture

That's a perfectly valid solution. Nice work!

I re-draw the triangles in a separate sheet and added the corresponding values as needed. Only was missing from the small triangle was to add the information that the 3rd angle was y and finally apply the rule. Right?
greenlight-admin's picture

That's correct.

So for this problem what I have found is that the comparison of triangles can only happen within a single triangle not for two separate triangles. In this problem since we know the triangles are similar can we then compare the angles of triangles of the two triangles?
greenlight-admin's picture

Sorry ravin654, but I'm not sure what you're asking. Can you rephrase your question?

So my question is the reason why you compared the two triangles to each other is because they are similar? Usually the property for longest side opposite biggest angle is only for comparing one triangle not two. But in this case you compared it for two triangles? Does this make sense now?
greenlight-admin's picture

Thanks for the clarification.
I believe you may be confusing the triangle property noted at 1:01 in this video with a different property regarding similar triangles covered here:

The first property allows us to the compare sides and angles in ONE triangle, and the second property allows us to compare sides in TWO similar triangles.

Does that help?

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