Question

Please guys someone help me with this geometry question :

The sides of quadrilateral ABCD are tangent to a circle (C) of center O . [ (C) is inscribed in the quadrilateral ABCD) .

-Show that AB + CD = AD + BC.

Answer 1

Hey CliffSedge :)

Answer 2

Howdy. I'm checking out your inscribed circle.

Answer 3

Oh alright, Cheers , thanks for trying to help.

Answer 4

I think you can make use of the equal radii of the circle, OE=OF=OG=OH.

Answer 5

Oh ya right, the radii are equal. We want to show that AB + CD =AD +BC . I thought about it i didn't know how to solve it

Answer 6

Oh, I got it! The tangents from a single point are equal.

Answer 7

So CD = CB and AD = AB?

Answer 8

No, but AH=AE, CG=CF, etc.

Answer 9

Yeah..

Answer 10

You can set up all those equality statements, then make the necessary substitutions.

Answer 11

BF = BE , AE = AH , DH = DG , CG = CF.

Answer 12

So how do i solve it? Didn't get how sorry.

Answer 13

You'll have to use a combination of segment addition postulate, maybe transitive property.
You essentially have to rearrange those equalities along with some addition statements until it looks like the statement you want to prove.

Answer 14

I'm trying to create the final form of the prove, I'm now knowing how.

Answer 15

Not*

Answer 16

This should get you started