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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 1203.
FEBM cyclic => CFM=CMF => FC=CMCFB+CMB=CFB+CFG => CMB=CFG=> MCAH=CFGDAt the same AF=AH etc
Let AD meet EC at NWe have AN ⊥NC and ∠ (NAC)= ∠ (CAN)= 45 degreesQua. ABCN is cyclic with AC as a diameterSo Bisector BF cut the midpoint N of arc ACAnd F, N and G are collinear and AF=FD and FC=EGAll sides and internal angles of quadrilateral AFGE congruent to DGFC .Qua. AHBF is cyclic and AB is an angle bisector of ∠ (HBF)=> AH=AF=DGSimilarly FC=CM=EGSo quadrilaterals AFGE , DGFC , AHMC are congruent ( corresponding sides and angles are congruent)
FCBM is concyclic, hence < HMC = CFG and further since BC bisects < FBM, FC= CMSo ∆ ACM ≡∆ FCD, SAS ...(1) and so < CFD = < AMCHence AM = FD, < AMH = < DFG < AHM = < DGF (since each angle = < CFB, AHBF being cycilc)So ∆ AHM ≡∆ FGD, ASA...(2)From (1) and (2) easily AHMC ≡GDFCWe can similarly show that AHMC≡AFGEHence AHMC ≡AFGE ≡GDFCSumith PeirisMoratuwaSri Lanka
Further conclusions from this problemEFCG and AFDG are parallelograms< HFM = 90
What are the conditions for 2 quadrilaterals to be congruent? (i.e. for all4 sides + 4 angles + 2 diagonals to be equal) 1. 4 sides + 1 diagonal equal (2 SSS triangles congruency) 2. 3 sides + 2 diagonals (2 SSS triangles congruency) 3. 4 sides + 1 corresponding included angle equal (1 SAS triangle and 1 SSS triangle congruency) 4. 3 sides + 2 corresponding included angles equal (so the diagonals are equal by SSS and the 4th side by SSS) 5. 2 sides + 3 corresponding angles equal (which of course makes the 4 th angle equal too + SAS congruency twice) Any more anyone?
From my solution 2 sides and 3 angles
In this case quadrilaterals are trapezoidsIt's enough: equal bases and altitude
Not sufficient - u can draw many different trapezoids with equal base and equal height
Just in this case "Right angle trapezoids"
Even then by changing the length of the side parellel to the base u can get many dufferent right angle trapezoids of equal base and equal height
Conclusion includes 2 bases
2 parallel sides + 2 right angles + height all equal Now the corresponding trapezoids are congruent
I think there is a ratio between area of EGBH and GDMB Depend it from point F (or from angles of ABC)