Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Thursday, March 31, 2016
Geometry Problem 1203: Right Triangle, Square, Angle Bisector, Three Congruent Quadrilaterals
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FEBM cyclic => CFM=CMF => FC=CM
ReplyDeleteCFB+CMB=CFB+CFG => CMB=CFG
=> MCAH=CFGD
At the same AF=AH etc
Let AD meet EC at N
ReplyDeleteWe have AN ⊥NC and ∠ (NAC)= ∠ (CAN)= 45 degrees
Qua. ABCN is cyclic with AC as a diameter
So Bisector BF cut the midpoint N of arc AC
And F, N and G are collinear and AF=FD and FC=EG
All sides and internal angles of quadrilateral AFGE congruent to DGFC .
Qua. AHBF is cyclic and AB is an angle bisector of ∠ (HBF)=> AH=AF=DG
Similarly FC=CM=EG
So quadrilaterals AFGE , DGFC , AHMC are congruent ( corresponding sides and angles are congruent)
ReplyDeleteFCBM is concyclic, hence < HMC = CFG and further since BC bisects < FBM, FC
= CM
So ∆ ACM ≡∆ FCD, SAS ...(1) and so < CFD = < AMC
Hence AM = FD,
< AMH = < DFG
< AHM = < DGF (since each angle = < CFB, AHBF being cycilc)
So ∆ AHM ≡∆ FGD, ASA...(2)
From (1) and (2) easily AHMC ≡GDFC
We can similarly show that AHMC≡AFGE
Hence AHMC ≡AFGE ≡GDFC
Sumith Peiris
Moratuwa
Sri Lanka
Further conclusions from this problem
ReplyDeleteEFCG and AFDG are parallelograms
< HFM = 90
What are the conditions for 2 quadrilaterals to be congruent? (i.e. for all
ReplyDelete4 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
ReplyDeleteIn this case quadrilaterals are trapezoids
ReplyDeleteIt's enough: equal bases and altitude
Not sufficient - u can draw many different trapezoids with equal base and equal height
DeleteJust in this case "Right angle trapezoids"
ReplyDeleteEven 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
DeleteConclusion includes 2 bases
ReplyDelete2 parallel sides + 2 right angles + height all equal
DeleteNow the corresponding trapezoids are congruent
I think there is a ratio between area of EGBH and GDMB
ReplyDeleteDepend it from point F (or from angles of ABC)