378. Some metrical relations.
by Virgil Nicula, Jun 11, 2013, 7:36 PM
PP1. Let 
-isosceles 
with the incenter 
and circumcenter 
. Let the lengths 
and 
of the the 
-bisector
and the-bisector respectively. Prove that 
, where 
is the midpoint of ![$[AB]$](//latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
.
Method 1. Let
where 
, 
on the 
-bisector ![$[AD]$](//latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
, where 
. Thus, 
, 
. Thus, 
is an isosceles trapezoid, where
. So 
, i.e. 
.
Thus,
, i.e. 
is a rhombus, i.e. .
Method 2. Let , where , the -bisector , where . Thus, 
. In conclusion, , what means that is a rhombus, i.e. .
Method 3 (metric). Denote-bisector 
, where and 
, 
. Thus, 
. In conclusion, , what means that is a rhombus, i.e. .
PP2. Prove that in any
have 
, where 
are the lengths of the
-altitude and the -bisector respectively. Particular case. If 
, then there is the chain of equivalencies 
.
Proof. Denote
so that 
and 
so that 
and 
. Since 
obtain that 
. Thus, 
. Observe that 
. Apply the theorem
of Sines in
. Using 
obtain that
^2=a^3c\iff$](//latex.artofproblemsolving.com/c/9/5/c952168d1685bbdc519588f046624081819ad15c.png)
.
PP3. In let 
so that 
and . Prove that 
, where 
are the lengths of the -median and the -bisector.
Proof. Denote the midpoint of the -bisector ![$[BD]$](//latex.artofproblemsolving.com/3/2/6/3261e689901bce018ecdef47b9bc60a78ead3746.png)
, where 
. Thus, 
and 
. Therefore, 
is an isosceles
trapezoid, i.e.
. In conclusion, 
.
PP4. Let
be a triangle and 
, 
be two points for which denote 
. Prove that :
is tangential 
.
If for denote the tangent points 
, 
, 
, 
, then exists 
, where 
.
Exists 
which is tangent to sidelines of 
and 
and in this case is an extangentially.
See here => http://jl.ayme.pagesperso-orange.fr/Docs/Urquhart.pdf "the most elementary theorem of Euclidean geometry".
Proof. Suppose that
is a tangential quadrilateral and denote the tangent points , , , . Therefore:
.
.
Let 
and apply Menelaos' 
and 
.
.
PP5. Let
, 
be exterior circles and let 
be a circle exterior tangent to circles 
, 
in 
, 
respectively. An exterior common tangent to the
circles , touches these in 
, 
respectively so that 
separates , . Prove that 
is cyclic and 
are concurrently.
Proof.
Therefore,
is a cyclical quadrilateral. Denote 
and apply
the Menelaus's theorem to the points
and 
.
PP6. Let a rectangle
and 
, 
so that 
is equilateral. Prove that ![$[ABE]=X\ ,\ [ADF]=Y\ ,\ [CEF]=Z\implies$](//latex.artofproblemsolving.com/9/8/5/985654b3d2e3b048db2ae26ed02b7d178a613167.png)
.
Proof. Denote
. Observe that 
. Suppose w.l.o.g. 
.
Thus,
, what is truly.
Extension. A rectangle , , . Then ![$:\ \left\{\begin{array}{c}
[ABE]=X\ ,\ [ADF]=Y\ ,\ [CEF]=Z\\\\
m\left(\widehat{AFE}\right)=\alpha\ ,\ m\left(\widehat{AEF}\right)=\beta\ ,\ m\left(\widehat{EAF}\right)=\gamma\end{array}\right\|\implies$](//latex.artofproblemsolving.com/d/e/b/debd058187ee9183a40640bb6b704fe97b459fc9.png)
Proof 1. Denote
and 
. Is well-known that 
![$\frac 1{4\cdot [AEF]}$](//latex.artofproblemsolving.com/a/d/b/adb0057b601465428bdff6018c19159d2a5a94c7.png)
, where ![$[AEF]$](//latex.artofproblemsolving.com/1/3/c/13c70c42a7bfa54d17fc9289b2134a7a1fd4abcf.png)
is the area of . Prove easily that 
. The relation 
becomes
![$bx\left[\left(a^2+y^2\right)-(ax+by)\right]+$](//latex.artofproblemsolving.com/f/1/4/f142de8107fb990f37b20e106e7c8ad534aa4dca.png)
![$ay\left[\left(b^2+x^2\right)-(ax+by)\right]=$](//latex.artofproblemsolving.com/4/4/d/44d72fd03ccc0a8351f97fa6284fe5d9b5b613e5.png)
, what is truly. In conclusion, is truly.
Remark. I used
in , where ![$S=[ABC]$](//latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
.Indeed, 
Proof 2 With same notations from the upper proof show that the identity using the following relations 
.
.
.
Proof 3. Let . Observe that 
and 
.
Therefore,
. Obtain that 
, what is truly.
PP7. Let and an interior point 
for which 
and 
. Prove that 
.
Proof. Denote
. Thus, 
. Apply the Menelaus' theorem to 
.
PP8. Let the-right-angled and its -excircle 
. Denote 
and
. Prove that 
.
Proof. Denote
. Thus, 
and 
, i.e. 
. Since
we get 
, i.e. 
and 
. So 
. Thus have 
.
PP9. Let an isosceles trapezoid with 
so that ![$[CD]$](//latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
is a diameter of its circumcircle 
.
For
so that 
separates 
and , let 
. Prove that 
is constant.
Proof 1 (metric). Let
. Thus, 
and 
. Therefore,
![$2R-[2d+(x+y)]\implies$](//latex.artofproblemsolving.com/2/8/f/28f0d43c34cd1ed5b6af9bcc56cb15df67828030.png)
because![$d[2(R-d)-(x+y)]=2Rd-2d^2-d(x+y)\stackrel{(1)}{=}$](//latex.artofproblemsolving.com/6/b/e/6be9ff389c81d1b9a5447f084deeb92d4925e176.png)
.
Proof 2. Let
. Thus, 
and 
![$2(R-d)-[h(\tan x+\cot x)-2d]\implies$](//latex.artofproblemsolving.com/9/7/7/97704d57f0cffa943ce04cccc3f507a0e4ce56ef.png)
.
In conclusion,
(constant).
Particular case.
with 
.
Proof 3 (Cristhian Centeno). . So
. Denote 
so that 
. Prove
easily that
and 
![$xy=d[2(R-d)-(x+y)]\stackrel{(2)}{\implies}$](//latex.artofproblemsolving.com/5/7/d/57d27fef46e0a1fbbe25f6de336ee7aefd7cb155.png)
(constant).
PP10. Let and 
. Consider and 
so that 
and 
are the -symmedians in 
and 
respectively.
Denote
. Prove that : 
![$\sqrt[3]{\frac {VC}{VM}\cdot\frac {WM}{WA}}\ ;\ PR\parallel AC\ ;\ \frac {VC}{VM}\cdot\frac {WA}{WM}\ge 4$](//latex.artofproblemsolving.com/3/d/6/3d6a128808cd1bf252ba199cd8b8380bb03457d1.png)
.
Proof. Denote
. Apply the Menelaus' theorem to
Thus, the relation
, what is evidently. Apply Ceva's theorem to 
and in the mentioned and respectively :
In conclusion,
![$\frac {MC}{MA}=\sqrt[3]{\frac {VC}{VM}\cdot\frac {WM}{WA}}\ .$](//latex.artofproblemsolving.com/6/c/8/6c805a761e90695cd55621bd3cc9b717a267760b.png)
Thus, 
, what is truly. Observe that 
.
PP11 (Edson Curahua). Let
be a semicircle with the diameter . For a point 
define
the semicircles
, 
with the diameters ![$[AC]\ ,$](//latex.artofproblemsolving.com/6/e/1/6e12d67d19cbed5cac0510150cbffc8f6fdc8ce9.png)
![$[CB]$](//latex.artofproblemsolving.com/f/0/a/f0a45a13c6123c357e7ffe7d05e50e68ad7d157a.png)
respectively so that 
.
Construct the circle
which is interior tangent to 
and is exterior tangent to 
and 
. Prove thst the distance 
.
Proof. Observe that
and 
. Apply Stewart's relation in 
Denote 
. If 
, then 
and the area
![$S=[PRQ]=\sqrt{s(s-a)(s-b)(s-c)}\implies$](//latex.artofproblemsolving.com/3/7/f/37f301ef3ebba210cd2a910eaa28f00c603f5574.png)
. In conclusion, the distance 
from the point to the line is given by the relation
, i.e. 
. Prove easily that 
and 
is 
.
PP12 (Ruben Auqui Caceres). Let
be a semicircle with the diameter and a point 
for which denote 
so that
and 
. Consider the incircle 
of 
and the circle 
which is interior tangent to 
in the point
and which is tangent to the rays
, 
in the points , 
respectively. Prove that 
.
Proof. Denote
and from the - right 
obtain that 
.
Prove easily that
and 
. The inradius of is given by the relation 
![$2\left[-(1+x)+\sqrt{2(1+x)}\right]+\sqrt{2(1+x)}=$](//latex.artofproblemsolving.com/b/d/a/bda73c38c8685e7feacbd63dfa12de4f0e66d482.png)
.
Therefore,
and 
. Denote 
. Thus, 
i.e. 
.
Therefore,
. Apply the Ptolemy's theorem to 
. In conclusion, the ratio .
PP13. Let a rectangle with 
, 
and the circle 
with the diameter ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
.
For
so that 
separates and let 
. Prove that 
.
Proof. Denote
. Particular case 
becomes 
.
PP14. Let
so that separates , and exist 
, where I denoted 
- an open interval. Prove that ![$[PAB]\cdot [PCD]=[PBC]\cdot [PAD]\iff (A,M,N,D)$](//latex.artofproblemsolving.com/b/6/0/b60995bb8eb89f032fd45b86ec33df8ca52070f9.png)
is an harmonical division, i.e. 
what means that , 
are harmonical conjugate w.r.t. , 
.
Proof. Let
and 
. Thus, ![$[PAB]\cdot [PCD]=[PBC]\cdot [PAD]$](//latex.artofproblemsolving.com/5/0/8/5089d131e819bfaf9cc623c42cb3d1c4d98cd441.png)
. But 
. Thus, 
.
PP15 (lemma). Let
be a convex quadrilateral for which denote 
. Consider 
, 
. Then 
.
Proof. Denote
, where 
. Using the well-known relations 
obtain equivalencies
.
PP16 (Ruben Auqui). Let a square
with 
and 
. For 
let the midpoint
of ![$[YC]$](//latex.artofproblemsolving.com/9/4/5/945f77a1cbd4eb5059d6dd2c05273628a3ff56ce.png)
and the intersection 
. Prove that 
is tangent to 
.
Proof. Denote
, where 
. Thus, 
and 
. Apply the Menelaus' theorem to
. Observe that is tangent to 
. From the relations 
obtain that 
.
PP17. Let and 
so that 
. For 
construct the squares 
and 
. Prove that 
,
Proof. Denote the point so that 
is a square and 
. Apply the generalized Pythagoras'
.
.
Prove easily that
. In conclusion, 
![$OX^2+OY^2=6+2[(\sin 2u-\cos 2v)+(\sin 2v-\cos 2u)]\implies$](//latex.artofproblemsolving.com/3/5/7/35720e5d91984c8d38edcb167496070cc3a1f954.png)
.
PP18. Let
be a convex quadrilateral with 
and 
. Prove that 
.
Proof 1 (metric). Denote
and 
. Observe that 
and 
. Using the well-known identity 
. In conclusion, 
.
Proof 2 (trigonometric). Observe that
.
PP19. Let and an interior point for which denote 
. Prove that ![$[ABC]\cdot [DPE]=[APC]\cdot [DBE]$](//latex.artofproblemsolving.com/d/2/6/d26ed0daa38eab7e4a1b945d09bbb51fa9316594.png)
, where ![$[XYZ]$](//latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
is the area of the 
.
Proof. Apply the Menelaus's theorem
![$\frac {[ABC]}{[DBE]}=\frac {[APC]}{[DPE]}\implies$](//latex.artofproblemsolving.com/8/c/d/8cd09270756a20c954aed11632f4d20c2f43a698.png)
.
PP20 (Miguel Ochoa Sanchez). Let with 
. Prove that the area 
of the triangle 
is given by the relation 
.
Proof.
is acute and 
. Thus, 
and 
. Therefore, 
. Since 
obtain from the relations and 
, i.e.
. In conclusion, using the well-known identity 
and the relations 
and 
obtain finally that
, i.e. the required relation . I used 
. Observe that
.
PP21 (Ruben Huillca). Let with circumcircle . The circle 
is tangent to 
, 
and is exterior tangent to . Prove that 
, where 
is the -excircle of .
Proof. Denote
. Thus, 
, i.e. 
and 
. Thus, 
.
PP22. Let be a circle with the diameter and 
be a point for which denote 
and 
. Prove that 
.
Proof 1. Prove easily that the ray
is the bisector of 
, i.e. 
.
Proof 2. Let
. Theorem of Sines in 
.So 
i.e. .
Extension. Let with the circumcircle and 
. Let and
so that
. Prove that 
.
PP23. Prove that in any triangle there is the following equivalence: 
.
Proof 1 (metric).
.
Proof 2 (trigonometric). 
.
Remark.![$a^2=4S\tan\frac A2+2aR\tan\frac A2\left[1-\cos (B-C)\right]\implies$](//latex.artofproblemsolving.com/a/2/f/a2f6477b82e4bfa00fe1044d04b1f8f615dacc3b.png)
.
-isosceles 
with the incenter 
and circumcenter 
. Let the lengths 
and 
of the the 
-bisectorand the
-bisector respectively. Prove that 
, where 
is the midpoint of ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
.Method 1. Let
where 
, 
on the 
-bisector ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
, where 
. Thus, 
, 
. Thus, 
is an isosceles trapezoid, where
. So 
, i.e. 
.Thus,
, i.e. 
is a rhombus, i.e. .Method 2. Let
, where , the -bisector , where . Thus, 
. In conclusion, , what means that is a rhombus, i.e. .Method 3 (metric). Denote
-bisector 
, where and 
, 
. Thus, 
. In conclusion, , what means that is a rhombus, i.e. .PP2. Prove that in any
have 
, where 
are the lengths of the-altitude and the -bisector respectively. Particular case. If 
, then there is the chain of equivalencies 
.Proof. Denote
so that 
and 
so that 
and 
. Since 
obtain that 
. Thus, 
. Observe that 
. Apply the theoremof Sines in
. Using 
obtain that 
^2=a^3c\iff$](http://latex.artofproblemsolving.com/c/9/5/c952168d1685bbdc519588f046624081819ad15c.png)
.PP3. In
let 
so that 
and . Prove that 
, where 
are the lengths of the -median and the -bisector.Proof. Denote the midpoint
of the -bisector ![$[BD]$](http://latex.artofproblemsolving.com/3/2/6/3261e689901bce018ecdef47b9bc60a78ead3746.png)
, where 
. Thus, 
and 
. Therefore, 
is an isoscelestrapezoid, i.e.
. In conclusion, 
.PP4. Let
be a triangle and 
, 
be two points for which denote 
. Prove that :
is tangential 
.
If for denote the tangent points 
, 
, 
, 
, then exists 
, where 
.
Exists 
which is tangent to sidelines of 
and 
and in this case is an extangentially.See here => http://jl.ayme.pagesperso-orange.fr/Docs/Urquhart.pdf "the most elementary theorem of Euclidean geometry".
Proof. Suppose that
is a tangential quadrilateral and denote the tangent points , , , . Therefore:
.
.
Let 
and apply Menelaos' 
and 
.
.PP5. Let
, 
be exterior circles and let 
be a circle exterior tangent to circles 
, 
in 
, 
respectively. An exterior common tangent to thecircles
, touches these in 
, 
respectively so that 
separates , . Prove that 
is cyclic and 
are concurrently.Proof.
Therefore,
is a cyclical quadrilateral. Denote 
and applythe Menelaus's theorem to the points
and 
. PP6. Let a rectangle
and 
, 
so that 
is equilateral. Prove that ![$[ABE]=X\ ,\ [ADF]=Y\ ,\ [CEF]=Z\implies$](http://latex.artofproblemsolving.com/9/8/5/985654b3d2e3b048db2ae26ed02b7d178a613167.png)
.Proof. Denote
. Observe that 
. Suppose w.l.o.g. 
.Thus,
, what is truly.Extension. A rectangle
, , . Then ![$:\ \left\{\begin{array}{c}
[ABE]=X\ ,\ [ADF]=Y\ ,\ [CEF]=Z\\\\
m\left(\widehat{AFE}\right)=\alpha\ ,\ m\left(\widehat{AEF}\right)=\beta\ ,\ m\left(\widehat{EAF}\right)=\gamma\end{array}\right\|\implies$](http://latex.artofproblemsolving.com/d/e/b/debd058187ee9183a40640bb6b704fe97b459fc9.png)
Proof 1. Denote
and 
. Is well-known that 
![$\frac 1{4\cdot [AEF]}$](http://latex.artofproblemsolving.com/a/d/b/adb0057b601465428bdff6018c19159d2a5a94c7.png)
, where ![$[AEF]$](http://latex.artofproblemsolving.com/1/3/c/13c70c42a7bfa54d17fc9289b2134a7a1fd4abcf.png)
is the area of . Prove easily that 
. The relation 
becomes
![$bx\left[\left(a^2+y^2\right)-(ax+by)\right]+$](http://latex.artofproblemsolving.com/f/1/4/f142de8107fb990f37b20e106e7c8ad534aa4dca.png)
![$ay\left[\left(b^2+x^2\right)-(ax+by)\right]=$](http://latex.artofproblemsolving.com/4/4/d/44d72fd03ccc0a8351f97fa6284fe5d9b5b613e5.png)
, what is truly. In conclusion, is truly.Remark. I used
in , where ![$S=[ABC]$](http://latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
.Indeed, 
Proof 2 With same notations from the upper proof show that the identity
using the following relations 
.
.
.Proof 3. Let
. Observe that 
and 
.Therefore,
. Obtain that 
, what is truly.PP7. Let
and an interior point 
for which 
and 
. Prove that 
.Proof. Denote
. Thus, 
. Apply the Menelaus' theorem to 
.PP8. Let the
-right-angled and its -excircle 
. Denote 
and
. Prove that 
.Proof. Denote
. Thus, 
and 
, i.e. 
. Since
we get 
, i.e. 
and 
. So 
. Thus have 
.PP9. Let an isosceles trapezoid
with 
so that ![$[CD]$](http://latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
is a diameter of its circumcircle 
.For
so that 
separates 
and , let 
. Prove that 
is constant. Proof 1 (metric). Let
. Thus, 
and 
. Therefore,
![$2R-[2d+(x+y)]\implies$](http://latex.artofproblemsolving.com/2/8/f/28f0d43c34cd1ed5b6af9bcc56cb15df67828030.png)
because
![$d[2(R-d)-(x+y)]=2Rd-2d^2-d(x+y)\stackrel{(1)}{=}$](http://latex.artofproblemsolving.com/6/b/e/6be9ff389c81d1b9a5447f084deeb92d4925e176.png)
.Proof 2. Let
. Thus, 
and 
![$2(R-d)-[h(\tan x+\cot x)-2d]\implies$](http://latex.artofproblemsolving.com/9/7/7/97704d57f0cffa943ce04cccc3f507a0e4ce56ef.png)
.In conclusion,
(constant).Particular case.
with 
.Proof 3 (Cristhian Centeno).
. So 
. Denote 
so that 
. Proveeasily that
and 
![$xy=d[2(R-d)-(x+y)]\stackrel{(2)}{\implies}$](http://latex.artofproblemsolving.com/5/7/d/57d27fef46e0a1fbbe25f6de336ee7aefd7cb155.png)
(constant).PP10. Let
and 
. Consider and 
so that 
and 
are the -symmedians in 
and 
respectively.Denote
. Prove that : 
![$\sqrt[3]{\frac {VC}{VM}\cdot\frac {WM}{WA}}\ ;\ PR\parallel AC\ ;\ \frac {VC}{VM}\cdot\frac {WA}{WM}\ge 4$](http://latex.artofproblemsolving.com/3/d/6/3d6a128808cd1bf252ba199cd8b8380bb03457d1.png)
.Proof. Denote
. Apply the Menelaus' theorem to 
Thus, the relation
, what is evidently. Apply Ceva's theorem to 
and in the mentioned and respectively :
In conclusion,
![$\frac {MC}{MA}=\sqrt[3]{\frac {VC}{VM}\cdot\frac {WM}{WA}}\ .$](http://latex.artofproblemsolving.com/6/c/8/6c805a761e90695cd55621bd3cc9b717a267760b.png)
Thus, 
, what is truly. Observe that 
.PP11 (Edson Curahua). Let
be a semicircle with the diameter . For a point 
definethe semicircles
, 
with the diameters ![$[AC]\ ,$](http://latex.artofproblemsolving.com/6/e/1/6e12d67d19cbed5cac0510150cbffc8f6fdc8ce9.png)
![$[CB]$](http://latex.artofproblemsolving.com/f/0/a/f0a45a13c6123c357e7ffe7d05e50e68ad7d157a.png)
respectively so that 
.Construct the circle
which is interior tangent to 
and is exterior tangent to 
and 
. Prove thst the distance 
.Proof. Observe that
and 
. Apply Stewart's relation in 
Denote 
. If 
, then 
and the area![$S=[PRQ]=\sqrt{s(s-a)(s-b)(s-c)}\implies$](http://latex.artofproblemsolving.com/3/7/f/37f301ef3ebba210cd2a910eaa28f00c603f5574.png)
. In conclusion, the distance 
from the point to the line is given by the relation
, i.e. 
. Prove easily that 
and 
is 
. PP12 (Ruben Auqui Caceres). Let
be a semicircle with the diameter and a point 
for which denote 
so that
and 
. Consider the incircle 
of 
and the circle 
which is interior tangent to 
in the point and which is tangent to the rays
, 
in the points , 
respectively. Prove that 
.Proof. Denote
and from the - right 
obtain that 
.Prove easily that
and 
. The inradius of is given by the relation 
![$2\left[-(1+x)+\sqrt{2(1+x)}\right]+\sqrt{2(1+x)}=$](http://latex.artofproblemsolving.com/b/d/a/bda73c38c8685e7feacbd63dfa12de4f0e66d482.png)
.Therefore,
and 
. Denote 
. Thus, 
i.e. 
.Therefore,
. Apply the Ptolemy's theorem to 
. In conclusion, the ratio .PP13. Let a rectangle
with 
, 
and the circle 
with the diameter ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
.For
so that 
separates and let 
. Prove that 
.Proof. Denote
. Particular case 
becomes 
.PP14. Let
so that separates , and exist 
, where I denoted 
- an open interval. Prove that ![$[PAB]\cdot [PCD]=[PBC]\cdot [PAD]\iff (A,M,N,D)$](http://latex.artofproblemsolving.com/b/6/0/b60995bb8eb89f032fd45b86ec33df8ca52070f9.png)
is an harmonical division, i.e. 
what means that , 
are harmonical conjugate w.r.t. , 
.Proof. Let
and 
. Thus, ![$[PAB]\cdot [PCD]=[PBC]\cdot [PAD]$](http://latex.artofproblemsolving.com/5/0/8/5089d131e819bfaf9cc623c42cb3d1c4d98cd441.png)
. But 
. Thus, 
.PP15 (lemma). Let
be a convex quadrilateral for which denote 
. Consider 
, 
. Then 
.Proof. Denote
, where 
. Using the well-known relations 
obtain equivalencies
. PP16 (Ruben Auqui). Let a square
with 
and 
. For 
let the midpoint of ![$[YC]$](http://latex.artofproblemsolving.com/9/4/5/945f77a1cbd4eb5059d6dd2c05273628a3ff56ce.png)
and the intersection 
. Prove that 
is tangent to 
.Proof. Denote
, where 
. Thus, 
and 
. Apply the Menelaus' theorem to
. Observe that is tangent to 
. From the relations 
obtain that 
.PP17. Let
and 
so that 
. For 
construct the squares 
and 
. Prove that 
,Proof. Denote the point
so that 
is a square and 
. Apply the generalized Pythagoras' 
.
.Prove easily that
. In conclusion, 
![$OX^2+OY^2=6+2[(\sin 2u-\cos 2v)+(\sin 2v-\cos 2u)]\implies$](http://latex.artofproblemsolving.com/3/5/7/35720e5d91984c8d38edcb167496070cc3a1f954.png)
.PP18. Let
be a convex quadrilateral with 
and 
. Prove that 
.Proof 1 (metric). Denote
and 
. Observe that 
and 
. Using the well-known identity 
. In conclusion, 
.Proof 2 (trigonometric). Observe that
.PP19. Let
and an interior point for which denote 
. Prove that ![$[ABC]\cdot [DPE]=[APC]\cdot [DBE]$](http://latex.artofproblemsolving.com/d/2/6/d26ed0daa38eab7e4a1b945d09bbb51fa9316594.png)
, where ![$[XYZ]$](http://latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
is the area of the 
.Proof. Apply the Menelaus's theorem
![$\frac {[ABC]}{[DBE]}=\frac {[APC]}{[DPE]}\implies$](http://latex.artofproblemsolving.com/8/c/d/8cd09270756a20c954aed11632f4d20c2f43a698.png)
.PP20 (Miguel Ochoa Sanchez). Let
with 
. Prove that the area 
of the triangle 
is given by the relation 
.Proof.
is acute and 
. Thus, 
and 
. Therefore, 
. Since 
obtain from the relations and 
, i.e.
. In conclusion, using the well-known identity 
and the relations 
and 
obtain finally that
, i.e. the required relation . I used 
. Observe that
.PP21 (Ruben Huillca). Let
with circumcircle . The circle 
is tangent to 
, 
and is exterior tangent to
. Prove that 
, where 
is the -excircle of .Proof. Denote
. Thus, 
, i.e. 
and 
. Thus, 
.PP22. Let
be a circle with the diameter and 
be a point for which denote 
and 
. Prove that 
.Proof 1. Prove easily that the ray
is the bisector of 
, i.e. 
.Proof 2. Let
. Theorem of Sines in 
.So 
i.e. .Extension. Let
with the circumcircle and 
. Let and so that
. Prove that 
.PP23. Prove that in any triangle
there is the following equivalence: 
.Proof 1 (metric).
.Proof 2 (trigonometric).
.Remark.
![$a^2=4S\tan\frac A2+2aR\tan\frac A2\left[1-\cos (B-C)\right]\implies$](http://latex.artofproblemsolving.com/a/2/f/a2f6477b82e4bfa00fe1044d04b1f8f615dacc3b.png)
.This post has been edited 262 times. Last edited by Virgil Nicula, Jul 7, 2016, 4:23 PM
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