Geometry – Pythagoras’ Theorem

Posted on May 6, 2022 by kevin

In $\triangle{ABC}$, $AM$ is the median on the side $BC$. Prove that $AB^2+AC^2=2(AM^2 + BM^2)$

For $\triangle{ABC}$, $O$ is an inner point, and $D$, $E$, $F$ are on $BC$, $CA$, $AB$ respectively, such that $OD\perp BC$, $OE\perp CA$, and $OF\perp AB$. Prove that $AF^2+BD^2+CE^2=BF^2+DC^2+AE^2$.

$P$ is an interior point of $\triangle{ABC}$, $P_1$, $P_2$, and $P_3$ are exterior points outside of $AB$, $BC$, and $CA$, respectively. $PP_1\perp AB$, $PP_2\perp BC$, $PP_3 \perp AC$, and $BP_1 = BP_2$, $CP_2=CP_3$. Prove that $AP_1=AP_3$.

In square $ABCD$, $M$ is the midpoint of $AD$ and $N$ is the midpoint of $MD$. Prove that $\angle{NBC}=2\angle{ABM}$.

In $\triangle{ABC}$, $\angle{A}=90^\circ$, $AB=AC$, $D$ is a point on $BC$. Prove that $BD^2+CD^2 = 2AD^2$.

In $\triangle{ABC}$, $\angle{C}=90^\circ$, $D$ is the midpoint of $AC$. Prove that $AB^2+3BC^2=4BD^2$.

In $\triangle{ABC}$, $\angle{C}=90^\circ$, $E$, $D$ are points on $AC$ and $BC$ respectively. Prove that $AD^2+BE^2=AB^2+DE^2$.

In $\triangle{ABC}$, $\angle{C}=90^\circ$, $D$ is the midpoint of $AB$, $E$, $F $are two points on $AC$ and $BC$ respectively, and $DE\perp DF$. Prove that $EF^2=AE^2+BF^2$.

Let $ABCD$ be a convex quadrilateral. Prove that $AC\perp BD$ if and only if $AB^2+CD^2=AD^2+BC^2$.

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Geometry – Sides and Angles of Triangles

Posted on April 22, 2022 by kevin
  1. As shown in the diagram below, in $\triangle{ABC}$, $\angle{B}\gt \angle{C}$, $AD$ is the bisector of the $\angle{BAC}$, $AE\perp BC$ at $E$. Prove that $\angle{DAE}=\dfrac{1}{2}(\angle{B}−\angle{C})$.

2. There are four points $A$, $B$, $C$, $D$ on the plane, such that any three points are not collinear. Prove that in the triangles $ABC$, $ABD$, $ACD$ and $BCD$ there is at least one triangle which has an interior angle not greater than $45^\circ$.

3. In $\triangle{ABC}$, $AB=AC$, $D$, $E$, $F$ are on $AB$, $BC$, $CA$, such that $DE= EF=FD$. Prove that $\angle{DEB}=\dfrac{1}{2}(\angle{ADF}+\angle{CFE})$.

4. In $\angle{ABC}$, $AC=BC$, $\angle{C}=20^\circ$, $M$ is on the side $AC$ and $N$ is on the side $BC$, such that $\angle{BAN}=50^\circ$, $\angle{ABM}=60^\circ$. Find $\angle{NMB}$ in degrees.

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Geometry Challenge – 12

Posted on April 4, 2022 by kevin

Acute $\triangle{ABC}$ is inscribed inside circle centered at $O$. $P$ is on $BC$ and $AP\perp BC$, and $\angle{ACB}>\angle{ABC}$. Prove the following:

  1. $\angle{BAC}+\angle{OBC}=90^\circ$
  2. $\angle{OAP}=\angle{ACB}-\angle{ABC}$
  3. If $\angle{ACB}-\angle{ABC}\ge 30^\circ$, and $MB=MC$, then $MP\gt CP$
  4. If $\angle{ACB}-\angle{ABC}\ge 30^\circ$, then $\angle{BAC}+\angle{POC}<90^\circ$

Click here for the solutions.

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Geometry Challenge – 11

Posted on March 21, 2022 by kevin

Let $D$ be an arbitrary point on the side $BC$ of a given triangle $ABC$ and let $E$ be the intersection of $AD$ and the second external common tangent of the incircles of triangles $ABD$ and ACD. As $D$ assumes all positions between $B$ and $C$, prove that the point $E$ traces an arc of a circle. Click here for the proof.

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Geometry Challenge – 10

Posted on March 14, 2022 by kevin

In the diagram below, two congruent semi-circles that are tangent to each other, are inscribed inside a bigger semi-circle with its diameter as 10 unit. The area of the shaded region can be expressed as $\dfrac{a}{b}\pi$, where $a$ and $b$ are co-prime. Find the value of $a+b$.

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