tag:blogger.com,1999:blog-6933544261975483399.post5152926566982573225..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1203: Right Triangle, Square, Angle Bisector, Three Congruent QuadrilateralsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-6933544261975483399.post-71946853577216902752016-04-03T08:10:14.914-07:002016-04-03T08:10:14.914-07:00I think there is a ratio between area of EGBH and ...I think there is a ratio between area of EGBH and GDMB <br />Depend it from point F (or from angles of ABC)c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-60045025431076682722016-04-03T05:12:01.162-07:002016-04-03T05:12:01.162-07:002 parallel sides + 2 right angles + height all eq...2 parallel sides + 2 right angles + height all equal <br /><br />Now the corresponding trapezoids are congruent Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-1124214109902467362016-04-03T01:24:40.211-07:002016-04-03T01:24:40.211-07:00Conclusion includes 2 basesConclusion includes 2 basesc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27656755172972934442016-04-02T19:10:34.055-07:002016-04-02T19:10:34.055-07:00Even then by changing the length of the side parel...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 Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-52751579236237345932016-04-02T10:30:59.267-07:002016-04-02T10:30:59.267-07:00Just in this case "Right angle trapezoids&quo...Just in this case "Right angle trapezoids"c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-80016884863925231002016-04-01T22:12:51.046-07:002016-04-01T22:12:51.046-07:00Not sufficient - u can draw many different trapezo...Not sufficient - u can draw many different trapezoids with equal base and equal height Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-86838917849765200042016-04-01T08:22:05.752-07:002016-04-01T08:22:05.752-07:00In this case quadrilaterals are trapezoids
It'...In this case quadrilaterals are trapezoids<br />It's enough: equal bases and altitudec.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65304659813760230262016-04-01T08:18:01.160-07:002016-04-01T08:18:01.160-07:00From my solution 2 sides and 3 anglesFrom my solution 2 sides and 3 anglesc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48517089428826620512016-04-01T04:38:09.571-07:002016-04-01T04:38:09.571-07:00What are the conditions for 2 quadrilaterals to be...What are the conditions for 2 quadrilaterals to be congruent? (i.e. for all<br />4 sides + 4 angles + 2 diagonals to be equal)<br /><br /><br /> 1. 4 sides + 1 diagonal equal (2 SSS triangles congruency)<br /><br /><br /> 2. 3 sides + 2 diagonals (2 SSS triangles congruency)<br /><br /><br /> 3. 4 sides + 1 corresponding included angle equal (1 SAS triangle and 1<br /> SSS triangle congruency)<br /><br /><br /> 4. 3 sides + 2 corresponding included angles equal (so the diagonals are<br /> equal by SSS and the 4th side by SSS)<br /><br /><br /> 5. 2 sides + 3 corresponding angles equal (which of course makes the 4<br /> th angle equal too + SAS congruency twice)<br /><br /><br /> Any more anyone?<br /><br /><br /> Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-20361703057712273332016-04-01T01:46:03.193-07:002016-04-01T01:46:03.193-07:00Further conclusions from this problem
EFCG and AF...Further conclusions from this problem<br /><br />EFCG and AFDG are parallelograms<br /><br />< HFM = 90Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-15352570965059800012016-04-01T01:36:54.647-07:002016-04-01T01:36:54.647-07:00FCBM is concyclic, hence < HMC = CFG and furthe...<br />FCBM is concyclic, hence < HMC = CFG and further since BC bisects < FBM, FC<br />= CM<br /><br />So ∆ ACM ≡∆ FCD, SAS ...(1) and so < CFD = < AMC<br /><br />Hence AM = FD,<br /> < AMH = < DFG<br /> < AHM = < DGF (since each angle = < CFB, AHBF being cycilc)<br /><br />So ∆ AHM ≡∆ FGD, ASA...(2)<br /><br />From (1) and (2) easily AHMC ≡GDFC<br /><br />We can similarly show that AHMC≡AFGE<br /><br />Hence AHMC ≡AFGE ≡GDFC<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-16053612802424855102016-03-31T21:10:21.942-07:002016-03-31T21:10:21.942-07:00Let AD meet EC at N
We have AN ⊥NC and ∠ (NAC)= ∠...Let AD meet EC at N<br />We have AN ⊥NC and ∠ (NAC)= ∠ (CAN)= 45 degrees<br />Qua. ABCN is cyclic with AC as a diameter<br />So Bisector BF cut the midpoint N of arc AC<br />And F, N and G are collinear and AF=FD and FC=EG<br />All sides and internal angles of quadrilateral AFGE congruent to DGFC .<br />Qua. AHBF is cyclic and AB is an angle bisector of ∠ (HBF)=> AH=AF=DG<br />Similarly FC=CM=EG<br />So quadrilaterals AFGE , DGFC , AHMC are congruent ( corresponding sides and angles are congruent)<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-46468328146603310152016-03-31T13:23:17.264-07:002016-03-31T13:23:17.264-07:00FEBM cyclic => CFM=CMF => FC=CM
CFB+CMB=CFB+...FEBM cyclic => CFM=CMF => FC=CM<br />CFB+CMB=CFB+CFG => CMB=CFG<br />=> MCAH=CFGD<br />At the same AF=AH etcc.t.e.onoreply@blogger.com