Circles and Triangle Geometry

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tempusedaxrerum

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tempusedaxrerum

#1 h Mar 11, 2017, 2:05 AM • 1 Y

Y by Adventure10

1. The triangle $ABC$ , where $AB<AC$ , has a circumcircle $S$ . The perpendicular from $A$ to $BC$ intersects $S$ again at $P$ . The point $X$ lies on the line segment $AC$ , and $BX$ intersects $S$ again at $Q$ . Show that $BX=CX \iff PQ$ is a diameter of $S$ .

2. Let $ABC$ be an isosceles triangle with $AB=AC$ and $\angle BAC=80^\circ.$ $O$ is a point inside the triangle such that $\angle OBC = 10 ^\circ$ and $\angle OCB = 20^\circ$ Find $\angle OAC$

3. Let $ABCD$ be a convex quadrilateral with $\angle ABC = \angle ADC$ . Suppose that $E, F, G, H$ are the midpoints of $AC, BD, AD, CD$ respectively.

(i) Prove that $EFGH$ is a cyclic quadrilateral
(ii) Hence prove that $\angle AEF = \angle ACB - \angle ACD$ .

4. Also, just a question, if a quadrilateral can be inscribed in a semi-circle with one side lying on the diameter does it imply it is cyclic? If yes is it possible to prove it?

Thanks in advance

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tempusedaxrerum

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tempusedaxrerum

#2 h Mar 11, 2017, 6:06 AM • 2 Y

Y by Adventure10, Mango247

any ideas guys? thanks

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AdBondEvent

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AdBondEvent

#3 h Mar 11, 2017, 6:23 AM • 2 Y

Y by Adventure10, Mango247

For the first , $BX = CX$ gives us $\angle{XBC} = \angle{XCB}$ and $\angle{PBC} = \angle{PAC}$ due to equal angles subtended by same segment on circle . But , $\angle{PAC} + \angle{ACB} = 90$ , therefore , $\angle{QBP} = \angle{QBC} + \angle{PBC} = 90$ . For the other part , if $PQ$ is diameter , then $\angle{PBC} + \angle{CBQ} = 90$ . But , $\angle{PBC} = \angle{PAC}$ . Therefore $\angle{CBQ} = 90 - \angle{PAC} = \angle{BCX}$ .

This post has been edited 1 time. Last edited by AdBondEvent, Mar 11, 2017, 6:29 AM

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Tumon2001

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Tumon2001

#4 h Mar 11, 2017, 7:11 AM • 2 Y

Y by Adventure10, Mango247

Solution ( $1$ ):

Apply power of a point to get $AX = QX$ .
This implies that $AQ || BC$ . So, $\angle PAQ$ = $90$ ° and the result follows.

The 'iff' part follows directly as $AQCB$ becomes an isosceles trapezium.

This post has been edited 2 times. Last edited by Tumon2001, Mar 11, 2017, 7:36 AM

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AdBondEvent

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AdBondEvent

#5 h Mar 11, 2017, 7:16 AM • 2 Y

Y by Adventure10, Mango247

Tumon2001 wrote:

Solution ( $1$ ):

Apply power of a point to get $AX = QX$ .
This implies that $AQ || BC$ . So, $\angle PAQ$ = $90$ ° and the result follows.

@above you would have to show the only if part also (Though its not hard) ...

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Tumon2001

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Tumon2001

#6 h Mar 11, 2017, 7:38 AM • 1 Y

Y by Adventure10

AdBondEvent wrote:

Tumon2001 wrote:

Solution ( $1$ ):

Apply power of a point to get $AX = QX$ .
This implies that $AQ || BC$ . So, $\angle PAQ$ = $90$ ° and the result follows.

@above you would have to show the only if part also (Though its not hard) ...

Oops! I didn't notice it.
Anyway, I have shown it now.

This post has been edited 2 times. Last edited by Tumon2001, Mar 11, 2017, 7:40 AM

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tempusedaxrerum

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tempusedaxrerum

#7 h Mar 11, 2017, 8:28 AM • 2 Y

Y by Adventure10, Mango247

Thanks guys! Any ideas (or hints) on the remaining qns?

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Tumon2001

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Tumon2001

#8 h Mar 11, 2017, 1:05 PM • 2 Y

Y by AlexLewandowski, Adventure10

Solution ( $3$ ):

( $i$ ): Let $\angle ABC = \angle ADC = x$ .
The mid-point theorem implies that $GF || AB$ , $GE || CD$ and $EH || AD$ .
So, $GEHD$ is a ||gm. Thus, $\angle GEH = x$ .
Also, $\angle GFH = \angle GFD + \angle HFD = \angle ABD = \angle CBD = x$ .
Hence, $\angle GEH = \angle GFH$ and the result follows.

( $ii$ ): This proof will vary with the diagram. In my diagram, $F$ lies on the same side of $AC$ as $D$ . In this case,
$\angle ACB = \angle ACD - \angle AEF$ .
To prove this, let $X$ be the midpoint of $AB$ .
The mid-point theorem implies that $FXEF$ is a ||gm. Also, $EX || BC$ .
So, $\angle ACB = \angle AEX = \angle FEX - \angle AEF = \angle EFH - \angle AEF = \angle EGH - \angle AEF = \angle ACD - \angle AEF$ .
This completes the proof.

This post has been edited 5 times. Last edited by Tumon2001, Mar 11, 2017, 1:11 PM

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svatejas

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svatejas

#9 h Mar 11, 2017, 1:17 PM • 1 Y

Y by Adventure10

If $\angle ABC=x,$ then how is $\angle ABD,CBD=x$ ?

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svatejas

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svatejas

#10 h Mar 11, 2017, 1:37 PM • 2 Y

Y by Adventure10, Mango247

2. Since $\angle BAC$ is the greatest in $\Delta ABC$ , $BC>AB=AC$ .

Take BD=AB=AC on side BC and rest is angle chasing. You get $\angle OAC=30^{\circ}$ .

This post has been edited 3 times. Last edited by svatejas, Mar 11, 2017, 1:40 PM

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svatejas

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svatejas

#11 h Mar 11, 2017, 1:49 PM • 2 Y

Y by Adventure10, Mango247

4.No. You can in fact prove that it won't be a cyclic quadrilateral in many ways.

Simplest way to prove is by considering a non isosceles trapezium(which is not cyclic) inscribed in the semi circle.

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sunken rock

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sunken rock

#12 h Mar 11, 2017, 6:37 PM • 2 Y

Y by Adventure10, Mango247

For #2:
Take $D$ isogonal conjugate of $O$ w.r.t. $\triangle ABC$ and $E$ the circumcenter of $\triangle BCD$ ; $\triangle BED$ is equilateral and $E$ lies onto the perpendicular bisector of $BC$ . Easy angle chase shows that, since $\angle ABE=70^\circ$ , then $\triangle ABE$ is isosceles, $70^\circ-40^\circ-70^\circ$ . Thus $AD$ is angle bisector of $\angle BAE$ and $\angle BAD=20^\circ$ ; as constructed, $\angle BAD=\angle CAO=20^\circ$ .

Remark: Taking $F$ as intersection of $AB$ with the perpendicular from $C$ to $BO$ gives, according to double angle lemma $\angle AFO=30^\circ$ , thus $FAOC$ is cyclic and subsequently $\angle CAO=\angle CFO=20^\circ$ .

Best regards,
sunken rock

This post has been edited 1 time. Last edited by sunken rock, Mar 12, 2017, 7:58 PM

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tempusedaxrerum

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tempusedaxrerum

#13 h Mar 12, 2017, 1:45 AM • 2 Y

Y by Adventure10, Mango247

Thanks for the quick responses guys! Really appreciate your help!

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