Geometry Q (1 Viewer)

Axio · May 9, 2014

Hey guys

So I am having trouble with part ii of this question. I can do everything including proving that AY = CX but I can't prove that BZ = AY =CX. I'd appreciate some help. Thanks.

View attachment 00.pdf

I97 · May 9, 2014

not 100% sure but if you prove triangle BCZ is congruent to triangle ACY, then BZ=AY and since AY=CX, BZ=AY

SpiralFlex · May 9, 2014

$a) \;$We want to prove that$\; \triangle BXC \equiv \triangle BAY$

${\color{Blue} \textup{Thought process:}}\;We are given that$\; \triangle BXA, \triangle AZC, \triangle BYC \;$are all equilateral which implies all the respective interior angles of the triangles listed at$ \;60^\circ. \;$It also implies that all the sides of each triangle are equal, I will use one mark, two mark, three marks to distinguish that the sides may have different lengths.$

$$If I mark in the diagram and focus on$\; \triangle BXC \;$and$\; \triangle BAY.\;$Can you see why$\; \angle XBA = YBA?, BX = BA, BC = BY.\; $Now why can we conclude from this$\; \triangle BXC \equiv \triangle BAY?$

$b)\; $What does this say about$\; CX \;$and$\; AY. \;$Why?$$

$$Now construct the line$\; BZ \;$and can you argue similarly for$\; \triangle BAZ \;$and$\; \triangle ACY? \;$(i.e$\; AY = BZ)$

$$So after you've done that, we have$\; CX = AY, AY = BZ \;$and equality is transitive, so$\; BZ=AY=CX \;$as required$.$

Note: That is my thought process on approaching the problem, you need not to write it down like this in the solution. I just thought I'd show my thinking rather than the solution (as it may seem I pulled it out from the air)

Axio · May 10, 2014

So triangles BZC and ACY are congruent due to SAS. Thanks

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SpiralFlex · May 10, 2014

Geometry Q (1 Viewer)

Axio

=o

I97

Active Member

SpiralFlex

Well-Known Member

Axio

=o

SpiralFlex

Well-Known Member

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