A modified geometry problem

Geometry Level 3

A B C ABC is a 3-4-5 right triangle. Its median C D CD and A E AE intersects at F F . F G FG and F H FH are perpendicular to A B AB and B C BC respectively. Then D G + E H = ? DG+EH=\ ?

Bonus: F G = ? FG=?


The answer is 1.166666666666667.

Fahim Muhtamim by Fahim Muhtamim , 17, Bangladesh

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1 solution

Since CD and AE are medians, the intersection cuts them by ratio 2:1. This means GF = BH = 2/3 * BE. So EH is 1/3rd of BE. This means EH is 1/3 * 2 = 2/3. In a similar way we find DG is 1/2. Thus the requested sum is 7/6 = 1.167.

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