 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Distance = Inradius (Posted on 2010-10-14) M is the midpoint of the side BC of triangle ABC, the center of whose incircle is denoted by O. AH is perpendicular to BC, and the line MO intersects AH at the point E. It is known that: |AB| ≠ |AC|.

Prove that the distance AE is equal to the radius of the incircle.

 No Solution Yet Submitted by K Sengupta Rating: 2.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 1 of 1
`Let H be the foot of the altitude fromvertex A to side BC and D the foot ofthe perpendicular from O to side BC. Let a = |BC|, b = |CA|, c = |AB|,h = |AH|, r = |OD| (the inradius),and s = (a+b+c)/2.`
`Assume b < c ( if b > c, then swap theletters B-C and b-c in the followingproof).`
`   |BM| = a/2   |BD| = s-b = (a-b+c)/2   |BH| = (a^2 - b^2 + c^2)/(2a)   |MD| = |BD|-|BM| = (c-b)/2   |MH| = |BH|-|BM| = (c-b)(c+b)/(2a)`
`From the similar right triangles MDOand MHE:`
`       r        |OD|     |EH|   --------- = ------ = ------    (c-b)/2     |MD|     |MH|`
`                     |EH|             = -----------------                (c-b)(c+b)/(2a)`
`             or`
`                r(b+c)        |EH| = --------                  a`
`Area of triangle ABC = rs = ah/2`
`             or`
`           h = 2rs/a`
`                            r(b+c)    |AE| = |AH|-|EH| = h - --------                                           a`
`            2rs     r(b+c)         = ----- - --------             a        a`
`            r                  r         = --- [2s - (b+c)] = --- [a]            a                  a`
`         = r`
` `

 Posted by Bractals on 2010-10-14 15:43:33 Please log in:

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