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**Given:** A circle inscribed in a right angled ΔABC. If ∠ACB = 90° and the radius of the circle is r.

**To prove:** 2r = a + b – c

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#### Solution

**Proof:** In given figure,

`{:("AF" = "AE"),("FB" = "BD"),("EC" = "DC"):}}` .....(i) [Tangent Segment theorem]

In ▢ODCE,

∠ECD = 90° ......[∠ACB = 90°, A–E–C, B-D–C]

`{:(∠"ODC" = 90^circ),(∠"OEC" = 90^circ):}}` ......[Tangent theorem]

∴ ∠EOD = 90° …[Ramining angle of ▢ODCE]

∴ ▢ODCE is a rectangle.

. Also, OE = OD = r ......[Radii of the same circle]

∴ ▢ODCE is a square ......`[("A Rectangle is square if it's"),("adjcent sides are congruent")]`

∴ OE = OD = CD = CE = r ......(ii) [sides of the square]

Consider R.H.S. = a + b – c

= BC + AC – AB

= (BD + DC) + (AE + EC) – (AF + FB) ......[B–D–C, A–E–C, A–F–B]

= (FB + r) + (AF + r) – (AF + FB) ......[From (i) and (ii)]

= FB + r + AF + r – AF - FB

= 2r

= L.H.S

∴ 2r = a + b – c

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