Geometry Test #?

Geometry Level 5

Given an triangle ABC. M is a midpoint of BC. H is the orthocenter. On the ray $\overrightarrow{HM}$ , draw a point K such that $MH=MK$ . Reflect the point H in BC and name this new point "I". BK meets HI at G.

What is the condition of triangle ABC so that GHCK is an isosceles trapezoid?

ABC is half of equilateral triangle. AB=AC ABC is an equilateral triangle. AB=BC AC=BC

\widehat{C}=90

degrees

\widehat{A}=90

degrees

\widehat{B}=90

degrees

1 solution

Alex Burgess
Jun 7, 2019

Rotating $ABC$ about $M$ we form triangle $CB'A$ .

Letting $CH$ hit $AB$ at $x$ and $AK$ hit $B'C$ at $x'$ , we can see $BxCx'$ forms a rectangle.

For $GHCK$ to be an isosceles trapezoid, we require $HG$ = $CK$ , so $AG - Hx = Kx' = Hx$ . Hence $AG = 2HX$ .

This occurs when $AB = 2xB$ , which is when $x$ is the midpoint of $AB$ and $AC = BC$ .

(Note: when $ABC$ is an isosceles triangle, $G=K$ and $GHC$ is an isosceles triangle too).

Note: $AB < AC$ is the condition that ensures $G$ is on the $A$ -side of $K$ .

Alex Burgess - 2 years ago

Can also be reasoned, without a rotation, like this:

Since M is the middle of HK, and also the middle of BC, BHCK is a parallelogram. Because H is the orthocentre, BH is at right angles to AC, and hence ACK is a right angle.

For GHCK to be an isosceles trapezoid, there are two conditions: Firstly, HC must be parallel to GK. This is already guaranteed by the parallelogram. Secondly, the angles HCK and GHC must be equal. If HCK = GHC = α, then ACH=90-α, and since CH is at right angles to AB, this means that CAB = α. Because BG is parallel to HC, GBA is a right angle, and also HGB=GHC, So HGB=α. The little rectangular triangle of which GB is the hypothenuse tells us that HGB+GBC=90, so that GBC=90-α. Then CBA=GBA-GBC=α. So, Triangle ABC has equal angles α at A and B, so it is an isosceles triangle with AC=BC.

K T - 1 year, 10 months ago

Geometry Test #?

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