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

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

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## Sunday, April 9, 2017

### Geometry Problem 1327: Two Squares Side by Side, Perpendicular, 90 Degrees, Angle

Labels:
angle,
geometry problem,
perpendicular,
square

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Problem 1327

ReplyDeleteIs triangle AED=triangle CED then AE=CG and <EAD=<DCG. But <DCG+<DGC=90.

So <DAE+<DGC=90.Therefore AE is perpendicular at CG.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

Problem 1327

ReplyDeleteIs triangle AED=triangle CED(AD=DC,DE=DG, <ADE=90=<CDG) then AE=CG and <EAD=<DCG. But <DCG+<DGC=90.

So <DAE+<DGC=90.Therefore AE is perpendicular at CG.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

Draw circles through ABCD and EFGD. L' intersection point.

ReplyDeleteang DL'G=45°, ang DL'C=135°=> L≡L' => ALC=90°

Tr.s ADE & CDG are congruent SAS

ReplyDeleteHence < DAE = < DCG

So ADLC is concyclic

Therefore < ALC = < ADC = 90

Sumith Peiris

Moratuwa

Sri Lanka

Rotate triangle CDG by 90 degs about D so that G goes to E, C to A, done.

ReplyDelete