Let us add some more points to our problem:
Let L, M, and N be the incenters and excenter of ΔABP,
ΔABC, and ΔACP respectively. Let the incircle of ΔABP
touch sides AP and BP at points D and F respectively. Let
the excircle of ΔACP touch rays PA and PC at points E
and G respectively.
Part (a)
|PE| = |PG| (1)
|PD| = |PF| (2)
|DE| = |FG| (1) minus (2)
|IA| + |AJ|
= |DA| + |AE|
= |DE|
= |FG|
= |FB| + |BC| + |CG|
= |IB| + |BC| + |CJ|
Part (b)
From similar right triangles LFB and BKM:
r1 |LF| |BK| |BK|
------ = ------ = ------ = ------
|BF| |FB| |KM| r
From similar right triangles NGC and CKM:
r2 |NG| |CK| |CK|
------ = ------ = ------ = ------
|CG| |GC| |KM| r
With the semiperimeter of ΔABC being
s = (|AB|+|BC|+|CA|)/2, lets calculate the
distances from vertices to points of tangency;
|BK| = s - |CA|,
|CK| = s - |AB|,
|BF| = (|AB|+|BP|-|PA|)/2,
|CG| = (|PA|+|AC|-|CP|)/2, and
|BF| + |CG| = s - |BC|.
Therefore,
r1|CK| + r2|BK|
= (|BF||BK|/r)*|CK| + (|CG||CK|/r)*|BK|
= (|BK||CK|/r)*[|BF| + |CG|]
(s-|CA|)(s-|AB|)
= ------------------ * (s-|BC|)
r
(s-|CA|)(s-|AB|)(s-|BC|)
= --------------------------
r
s(s-|CA|)(s-|AB|)(s-|BC|)
= ---------------------------
r*s
Area(ΔABC)2
= -------------
Area(ΔABC)
= Area(ΔABC)
QED
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