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

Labels:
angle bisector,
congruence,
perpendicular,
right triangle

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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)