Let-denote-the-circumcircle-of-ABC-The-tangent-to-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-

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 .