Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.

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

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## Monday, March 23, 2020

### Geometry Problem 1465: Tangential Quadrilateral, Incenter, Inscribed Circle, Equal Sum of Areas

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P1465

ReplyDeleteAE=AH, BE=BF, CG=FC, DG=DH => AE+BE+CG+DG=AH+BF+FC+DH =>

(AE+BE+CG+DG).R =(AH+BF+FC+DH).R

S yellow = S green

By tangent property, AH=AE, BE=BF, CF=CG, DG=DH

ReplyDeleteThus AD+BC=AB+CD

On the other hand, the height of those triangles are the same, namely the radius,

AD*r/2 + BC*r/2 = AB*r/2 + CD*r/2

Hence, S1+S3 = S2+S4.

We have 4 pairs of congruent Tr.s, (APH,APE), (BPE,BPF), (CPF,CPG), (DPH,DPG) one each of which is yellow and the other green.

ReplyDeleteThe result is thus easily derived

Sumith Peiris

Moratuwa

Sri Lanka