Let Ω denote the circumcircle of ABC. The tangent to Ω at A meets BC at X. Let the angle bisectors of ∠AXB meet AC and AB at E and F respectively. D is the foot of the angle bisector from ∠BAC on BC. Let AD intersect EF at K and Ω again at L(other than A). Prove that AEDF is a rhombus and further prove that the circle defined by triangle KLX passes through the midpoint of line segment BC.


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Question Number 112266 by Aina Samuel Temidayo last updated on 07/Sep/20

$Let Ω denote the circumcircle of ABC .$ $The tangent to Ω at A meets BC at X .$ $Let the angle bisectors of ∠ AXB meet$ $AC and AB at E and F$ $respectively . D is the foot of the angle$ $bisector from ∠ BAC on BC . Let AD$ $intersect EF at K and Ω again at$ $L (other than A) . Prove that AEDF is$ $a rhombus and further prove that the$ $circle defined by triangle KLX passes$ $through the midpoint of line segment$ $BC .$
	Answered by 1549442205PVT last updated on 07/Sep/20

	Commented by 1549442205PVT last updated on 07/Sep/20

	$i) Denote by H the midpoint of BC . Since$ $AL is the bisector ray of the angle \hat{A},$ $L is middle point of the smaller arc BC$ $Hence, three points O, H, L colinear .$ $and \hat{OHX} = 90 ° (property of the diameter)$ $On the other hands, \hat{OAX} = 90 ° (since$ $XA is the tangent at A of the circle (O))$ $We infer the quadrilateral AXHO is$ $inscribed) \Rightarrow \hat{AXH} + \hat{AOH} = 180 ° (1) . On the$ $other hands, the triangle OAL is isosceles$ $at O, so \hat{OAL} = \hat{OLA} = α . Moreover,$ $in the triangle OAL we have$ $\hat{OAL} + \hat{OLA} + \hat{AOH} = 180 ° infer$ $2 α + \hat{AOH} = 180 ° (2) . From (1) (2) we get$ $\hat{AXH} = 2 α . From that since \hat{AXE} =$ $\frac{1}{2} \hat{AXH} we infer \hat{OAL} = \hat{AXE} = α .$ $This gives us that \hat{KAX} + \hat{AXE} =$ $\hat{KAX} + \hat{OAL} = \hat{OAX} = 90 °$ $Thus, KD ⊥ EF . Consider two teiangles$ $XAC and XBA have the common angle$ $\hat{AXC}, \hat{ACB} = \hat{BAX} . It follows that$ $Δ AXC ∽ Δ BXA (a . a) \Rightarrow \frac{AX}{BX} = \frac{AC}{AB} (3)$ $. On the other hands, since AD is the$ $bisector of the angle \hat{A} we have$ $\frac{BD}{DC} = \frac{AB}{AC} (4) . Since XF is the bisector$ $of the \hat{AXB}, \frac{BF}{FA} = \frac{BX}{AX} (5) . From (3) (4)$ $(5) we obtain \frac{BD}{DC} = \frac{BF}{FA} \Rightarrow DF / / AC .$ $Similarly, we have DE / / AB, so$ $quadrilateral AEDF is a parallelogram$ $that has two mutual perpendicular$ $diagonals . That shows AEDF is a rhombus (q . e . d)$ $ii) By above proof we have O, H, L are$ $colinear (H - midpoint of BC) and$ $\hat{OLA} = \frac{1}{2} \hat{AXH} = \hat{KXH} = α . Therefore,$ $quadrilateral HKXL is inscribed which$ $shows that the circumcircle of triangle$ $KXL go through the midpoint H of BC$ $(q . e . d)$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$Thanks .$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$I really appreciate your efforts .$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$Are you there ?$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$I {don}^{'} t know if you can also help with$ $this please ?$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$Can you please show how DE / / AB ?$
	Commented by 1549442205PVT last updated on 07/Sep/20

	$because AL is bisector of \hat{BAC} .$
	Commented by 1549442205PVT last updated on 07/Sep/20

	$Prove \frac{CE}{EA} = \frac{CD}{DB} base on the hypothesis$ $XE is bisector \Rightarrow \frac{CE}{EA} = \frac{XC}{XA}, Δ XAC ∽$ $Δ XBA \Rightarrow \frac{XC}{XA} = \frac{AC}{AB} but \frac{AC}{AB} = \frac{CD}{DB} (as AD is bisector)$ $\Rightarrow \frac{CE}{EA} = \frac{CD}{DB} (q . e \sqrt{d)}$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$Thanks .$
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