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AE and BD are the angle bisectors of $\triangle ABC$ with in-center I. P is any point on DE produced. X, Y, and Z are the feet of the perpendiculars from P to CB, BA, and AC respectively.

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(1) Through P, draw a line parallel to AB cutting BC at $B_1$, BD at B’, AE at A’, and AC at $A_2$.

(2) Through B’, draw a line parallel to BC cutting AB at $B_2$, and AC at $C_1$. Let $PC_1$ cut BC at $C_2$.

CI, the third angle bisector of $\triangle ABC$, cuts $C_1C_2$ at F, cut $B_2C_1$ at C’, and cut PX produced at G. Then, $\triangle A’B’C’$ is a miniature of $\triangle ABC$ and with its in-center also located at I. Let C’A’ produced cut AB at $A_1$.

The question is:- “will $CC_1C’C_2$ be also a rhombus as the other two?”


A solution has been found through an alternate construction method. See question no. 2882211

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Yes, since it is also a parallelogram in which one diagonal is also an angle bisector, it will be a rhombus.

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  • $\begingroup$ I think we need to prove that $A'C'C_2$ is a straight line first before we can say it is a parallelogram. $\endgroup$ – Mick Aug 20 '18 at 8:48
  • $\begingroup$ Oh point! I didn't read the question properly $\endgroup$ – Sharv Laad Aug 20 '18 at 18:04

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