MHB Construct vertex D of an acute angled triangle

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To construct vertex D of an acute-angled triangle, the triangles must share the same base and lie between the same pair of parallel lines. The condition that CA equals CD is crucial; therefore, a line should be drawn through point C parallel to AB. This ensures that both triangles maintain the same height, as their areas are determined by the formula (1/2) base times height. By using a compass to draw an arc from point C with a radius equal to CA, point D can be accurately located where the arc intersects the parallel line. This method effectively satisfies all the geometric requirements outlined in the problem.
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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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