MHB Construct vertex D of an acute angled triangle

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SUMMARY

The discussion focuses on constructing vertex D of an acute-angled triangle, ensuring that segments CA and CD are equal. Participants emphasize the importance of drawing a line through point C that is parallel to line AB, as both triangles share the same base and must maintain equal heights. The area of the triangle is derived from the formula "(1/2) base times height," which underpins the geometric relationships discussed. The use of compasses to locate point D by intersecting an arc with the parallel line is confirmed as a correct approach.

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Any Ideas on how to begin?

Many Thanks :)
 
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You know the formula for the area of a triangle. Why don't you describe your ideas on the location of $D$?
 
:) Both the triangles should be on the same base and between same pair of parallel lines

I am not sure whether it correct but here are my ideas

egwh05.jpg


I think It has got to do something with parallelograms, I guess?

Many Thanks :)
 
mathlearn said:
Both the triangles should be on the same base and between same pair of parallel lines
Yes, but the problem statement also stipulates that $CA=CD$. So you should draw the line through $C$ that is parallel to $AB$ and then mark $D$ on that line so that $CA=CD$. To draw a parallel line through $C$, see here.
 
It has to do with the fact that the area of a triangle is "(1/2) base times height". Since the two triangles will have the same base, their heights must also be the same. That is the reason for drawing the line, through C, parallel to AB. The further condition is that "CA= CD". Set one leg of a pair of compasses at C, set the other on A, and draw an arc with center at C and radius CA. D is where the line and arc intersect.
 
Correct ?

vdjvyo.jpg
 
Yes, it's correct.
 

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