this is true for any right angled triangle i.e in a right angle triangle if a perpendicular is drawn on the hypotenuse from the opposite vertex, resulting two triangles will be similar to each other as well as the main triangle.
triangle drawn on semicircle on semicircle with diameter as hypotenuse is always a right angled triangle ,so the mention property of triangle will hold good.

in a triangle inside a triangle with any two vertices common the smaller trianglr will be similar to each other and to the bigger triangle

IMO D

scoowhoop wrote:See Attached.

Question: Are the smaller two triangles both proportional to the large triangle (XYZ)? If so will this always hold true for a triangle inscribed in a circle(where the hypotenuse is equal to the dia)?

In other words, will XY/q = YZ/4 and so on?

Attachment is not opening and turning me fanatical

I love this forum, thank you for all of the answers!

I'll try and attach the picture again.

Must be D.

I'll name the perpendicular vertex as "W".

You could just rephrase the question as - "What is the diameter?" OR "What are XW and WZ?"

Given: WY = 4

1. q = 2.

XYZ is a right triangle. And so is XYW and WYZ.

So, XW/YW is proportional to YW/WZ

XW=2, YW=4

2/4 = 4/WZ. WZ=8. You know what are XW and WZ. Sufficient.

2. r = 8

WZ=8
XW/YW is proportional to YW/WZ

XW/4 = 4/8

XW=2. You know what are XW and WZ. Sufficient.

scoowhoop wrote:See Attached.

Question: Are the smaller two triangles both proportional to the large triangle (XYZ)? If so will this always hold true for a triangle inscribed in a circle(where the hypotenuse is equal to the dia)?

In other words, will XY/q = YZ/4 and so on?

Your reservation that "will XY/q = YZ/4 and so on?" is right in the given state of affairs and it will hold true in all such applications.

The arc length in question is π times the radius = π × ½ (q + r).

Let's name the foot of perpendicular from Y on XZ as W, then...

∆XYZ ~ ∆XWY ~ ∆YWZ, such that

XW/WY = WY/WZ

Or q/4 = 4/r

Or q r = 16; and we want q + r.

(1) If q = 2, r is accessible and so is π × ½ (q + r). Sufficient


(2) If r = 8, q is accessible and so is π × ½ (q + r). Sufficient

[spoiler]D[/spoiler]