In a triangle $ABC$, $\angle B=90^o$ and $\angle A=60^o$, $I$ is the point of intersection of its angle bisectors. A line passing through the point $I$ parallel to the line $AC$, intersects the sides $AB$ and $BC$ at the points $P$ and $T$ respectively. Prove that $3PI+IT=AC$ . (Anton Trygub)
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Tags: geometry, incenter, right triangle
15.02.2021 00:16
@OP Are you sure this question is correct. I am getting that $AC=\sqrt{3}PT$, so obviously $AC<3PT$. I also made this geogebra sketch, which disproves the stated problem https://www.geogebra.org/calculator/hhvabw63
15.02.2021 00:22
incorrect original post parmenides51 wrote: In a triangle $ABC$, $\angle B=90^o$ and $\angle C=60^o$, $I$ is the point of intersection of its angle bisectors. A line passing through the point $I$ parallel to the line $AC$, intersects the sides $AB$ and $BC$ at the points $P$ and $T$ respectively. Prove that $3PT+IT=AC$ . (Anton Trygub) oops I had 2 typos (following the red ones) m the correct is parmenides51 wrote: In a triangle $ABC$, $\angle B=90^o$ and $\angle A=60^o$, $I$ is the point of intersection of its angle bisectors. A line passing through the point $I$ parallel to the line $AC$, intersects the sides $AB$ and $BC$ at the points $P$ and $T$ respectively. Prove that $3PI+IT=AC$ . (Anton Trygub) thanks for noticing