Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Sunday, April 30, 2017

### Geometry Problem 1334. Square, Kite, Sum of Segments

Labels:
kite,
quadrilateral,
square,
sum of segments

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Problem 1334

ReplyDeleteIs <GBE=<EBC=<EHG then GH=BG. Draw BK perpendicular in BE at point B (K belongs to ΑD).But triangle KBA=triangle BCE (rectangular, AB=BC,<KBA=<EBC) so KA=CE.Now rectangular triangle KBH is <KBG+<GBH=90=<BKH+<BHK so <KBG=<BKG ,then KG=BG=GH. Therefore GH=KG=KA+AG=CF+AG.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

∆EDH ~ ∆BCE => x²= ab + bc ( x = AD )

ReplyDeleteFrom ∆ABG => x² = c² - a²

From two rows c = a + b

Excellent solution

DeleteCan you please explain the first line?

DeleteEC/BC=DE/DH, EC=b, BC=x, DE=x-b, DH=c-(x-a)

DeleteExtend DA to X such that AX = b.

ReplyDeleteThen Tr.s BCE,BEF & AXB are all congruent SAS and so < ABX = < GBH = < CBE < GHB.

Since AB is perpendicular to XH, it follows that < XBH = 90

But GB = GH so each of these = XG

Therefore a+b =c

Sumith Peiris

Moratuwa

Sri Lanka

Solution 2

ReplyDeleteLet AB = p

From similar triangles b/p = p/(a+c)

So p^2 = b(a+c) = c^2 - a^2 and hence dividing by a+c, b = c-a

Sumith Peiris

Moratuwa

Sri Lanka

Further result < AFC = 135

ReplyDeleteDraw BK ⊥ BE, K on DA. Draw circle through H, B, K

ReplyDeleteG center of the cirle

Triangle GBH is Isosceles and GB=a+b (One way to derive is considering the right triangle BFE and applying pythogorus)

ReplyDelete=> c=a+b

Let length of square be 1.

ReplyDelete<CBE=<EHD

<CBE= tan(-1)b

<EHD= tan(-1){1÷(a+c)}

So, b(a+c)=1

BG=GH

SQRT(1+a^2)=c (∆BAG)

By substituting b(a+c)=1 into SQRT(1+a^2)=c, you get a+b=c.