## Thursday, March 31, 2016

### Geometry Problem 1203: Right Triangle, Square, Angle Bisector, Three Congruent Quadrilaterals

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.

1. FEBM cyclic => CFM=CMF => FC=CM
CFB+CMB=CFB+CFG => CMB=CFG
=> MCAH=CFGD
At the same AF=AH etc

2. Let AD meet EC at N
We 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)

3. FCBM 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

4. Further conclusions from this problem

EFCG and AFDG are parallelograms

< HFM = 90

5. What are the conditions for 2 quadrilaterals to be congruent? (i.e. for all
4 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?

6. From my solution 2 sides and 3 angles

7. In this case quadrilaterals are trapezoids
It's enough: equal bases and altitude

1. Not sufficient - u can draw many different trapezoids with equal base and equal height

8. Just in this case "Right angle trapezoids"

1. 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

9. Conclusion includes 2 bases

1. 2 parallel sides + 2 right angles + height all equal

Now the corresponding trapezoids are congruent

10. I think there is a ratio between area of EGBH and GDMB
Depend it from point F (or from angles of ABC)