Guess what? I will do class 8 geos ..

Obviously, I have become very rusted. So to unrust me ( oh god, is it unrust ?  I have become so bad in English). Okie wait.. "polish myself." I tried problems from ABJTOG 1.4. Easy ones TBH. Without further ado, here are the problems and solutions I tried. You guys can try too! I can assure you the difficulty is less than class 8. ( or class 7). I didn't try harder problems, because that would take me sometime. I did these in break :P . So yeah.. sorry for so easy  levels.

Milk and Mocha 

Problem 2 of ABJTOG:- Let $ABC$ be a triangle and let $M$ be a point on the ray $AB$ beyond $B$ such that $\overline{BM} = \overline{BC}$. Prove that $MC$ is parallel to the angle bisector of $\angle ABC$.

Solution :- Note that

$$\angle BMC=\frac{1}{2} \cdot (180-B)= \frac{B}{2}=\frac{1}{2}\angle ABC.$$

Problem 1 of ABJTOG :- Let $C$ be a point on the line segment $AB$. Let $D$ be a point that doesn’t lie on the line $AB$. Let $M$ and $N$ be points on the angle bisectors of $\angle ACD$ and $\angle BCD$, respectively, such that $MN \parallel AB$. Prove that the line $CD$ bisects $MN$.

Solution:- Clearly $\angle MCN=90.$ Now, to show that line $CD$ bisects $MN$, it's enough to show that $X$ is the centre of the $\Delta CMN ,$ where $X=CD\cap MN$ or it's enough to show that $\Delta MCX,\Delta CXN$ is isosceles, which follows from the parallel line and angle bisector stuff[ $\angle XMC= \angle MCA=\angle MCX.$

Problem 14 of ABJTOG:- Let $O, I, H$ be the circumcenter, incenter and orthocenter, respectively, of $|Delta ABC.$ Prove that $B, C, O, I, H$ lie on a circle if and only if $\angle BAC = 60$

Solution:- Since $B,C,I,H$ lie on one circle, we get

$$\angle BHC=BIC\implies 180-A=90+\frac A2\implies A=60.$$ So we get $\angle BHC=\angle BIC=120.$ But we also have $\angle BOC=120.$ Hence Prove that $B, C, O, I, H$ are cyclic

Problem 17 of ABJTOG:- Let $ABCDEF$ be a convex hexagon with $\overline{AB} = \overline{AF}, \overline{BC} = \overline{CD}$ and $\overline{DE} = \overline{EF}$. Prove that the angle bisectors of $\angle BAF, \angle BCD$ and $\angle DEF$ are concurrent.

Solution:- Well my one was exactly Thermos. So..

Note that the angle bisectors of $\angle BAF, \angle BCD$ and $\angle DEF$ are the perpendicular bisectors of $\overline{FB}, \overline{BD}, \overline{DF}$. Thus, they are concurrent at the circumcenter of $\triangle BDF$.

Problem 15 of ABJTOG:- Let $H$ and $O$ be the orthocenter and circumcenter in a triangle $ABC$, respectively. If $\angle BAC = 60$ , prove that $AH = AO$. Is the converse true?

Solution:- Just use the fact that $AH=2R\cdot \cos A.$ And $\cos A=\frac 12 \iff\angle A=60.$

Problem 40 of ABJTOG:- Let $D, E$ and $F$ be points on the sides $BC, CA$ and $AB,$ respectively, such that $BCEF$ is a cyclic quadrilateral. Let $P$ be the second intersection of the circumcircles of $\Delta BDF$ and $\Delta CDE.$ Prove that $A, D$ and $P$ are collinear.

Solution:- We apply radical axis theorem in circles $(BCEF),(BDF),(CDE).$ Note that $AB$ is the radical axis of $(BDF),(BFEC)$ and $AC$ is the radical axis of $(CDE),(BFEC).$ Since $AC,AB$ concur at $A$ and $DP$ is the radical axis of $(BDF),(CDE).$ Hence $A,D,P$ are collinear.

Problem 37 of ABJTOG:- Let $ABCD$ be a cyclic quadrilateral. The rays $AB$ and $DC$ intersect at $P$ and the rays $AD$ and $BC$ intersect at $Q.$ The circumcircles of $\Delta BCP$ and $\Delta CDQ$ intersect at $R.$ Prove that the points $P , Q$ and $R$ are collinear.

Solution:- Note that $\angle PBC=180-\angle PRC$ and $\angle PBC=\angle QDC=\angle QRC.$ Hence $\angle PRC+\angle QRC=180.$

Problem 38 of ABJTOG:- The diagonals of a cyclic quadrilateral $ABCD$ intersect at $S$. The circumcircle of $\Delta ABS$ intersects line $BC$ at $M$ , and the circumcircle of $ADS$ intersects line $CD$ at $N .$ Prove that $S, M$ and $N$ are collinear.

Solution:- Note that $\angle ABM= 180-\angle ASM$ and $\angle ABM=\angle ADN=\angle ASN.$ Hence $\angle ASN+\angle ASM=180.$ So $S, M$ and $N$ are collinear.

Problem 41 of ABJTOG :- Two circles are tangent to each other internally at a point $T$ . Let the chord $AB$ of the larger circle be tangent to the smaller circle at a point $P .$ Prove that $TP$ is the internal angle bisector of $\angle ATB.$

Solution:-Simple homothety.

Let $Z:= TP \cap (ATB).$ Note that $P$ is the lowest point, so by homothety, $Z$ will also be the "lowest" point i.e $Z$ is the midpoint of arc $AB.$ Hence $TZ=TP$ is the internal angle bisector of $\angle ATB.$

Problem 42 of ABJTOG:- Let $ABCD$ be a trapezoid $(AB || CD).$ Let $AC \cap BD = E$ and $AD \cap BC = F .$ Let $M, N$ be midpoints of $AB, CD,$ respectively. Prove that the points $E, F, M, N$ are collinear.

Solution:- Well clearly, $F,M,N$ are collinear. So, we will show that $E,M,N$ is collinear. Let $M'= EN\cap AB.$ By the parallel property, we get $\Delta  M'EB \sim NED$ with ratio $M'E/NE.$ So $M'B=ND \cdot \frac{M'E}{EN}.$ And we also have $\Delta  M'EA \sim NEC$ with ratio $M'E/NE .$ So $M'A=NC\cdot \frac{M'E}{EN}.\implies M'A=M'B\implies M=M'.$

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Well yeah.. that's the problems I did. Trust me the number of Olympiad problems, I am doing has exponentially decreased :(. 

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Sunaina 💜