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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1198.
Observe that triangle AEF congruent to tri. ABC ….( case ASA)So AF= AC= AGSo S(AFDC)= S(ABDE)= AB^2We have S(AHKG)=AG.AH=AC.AH= AB^2.. ( relation in right triangle ABC)So S(AFDC)=S(AHKG)
S(AHKG)= 2×S(AHG)= 2×S(ABG)= 2×S(AEC)= 2×S(AEB)= S(AEDB)= S(AFDB) + S(AEF)= S(AFDB) + S(ABC)= S(AFDC)
Since <EAF = < BAC, triangles AEF and ABC are congruent ASA and so AC = AFHence S(AFDC) = S(ABDE) = AB^2 = AH.AC since triangles ABH and ABC are similarNow AH.AC = AH.AG (since AC = AF = AG) = S(AHKG)So S(AFDC)= S(AHKG)Sumith PeirisMoratuwaSri Lanka
Angles FAE and CAB are equal, as angles with orthogonal hands. So, the triangles AEF and ABC are congruent; both have the same area. Thus, area of the square ABDE is equal to area ABDF plus area ABC, that is area(ACBDF) = area(ABDE) = c^2, where the length of side AB = c. On the other side, area(AGKH) = bp, where p = length of AH. However, in the right triangle ABC is valid c^2 = pb. It follows from the similar triangles ABC and AHB and c:p = b:c, that is c^2 = bp. Accordingly, area(AGKH) = area(ACDF).