VeritasKarishma wrote:
Kritisood wrote:
Responding to a pm:
Quote:
Can you explain why it's not BCA is similar to DCA? Since according to the image, A is the same for both triangles and C is the same, so shouldn't it be BCA and DCA?
Note that angles BCA and DCA are both 90 degrees so that's fine but angles BAC and DAC are not equal. These are two distinct angles and can take any values. Using the logic I have discussed in my solution, we find that triangle BCA is similar to triangle ACD.
Hi
VeritasKarishma. I understood that BAC and DAC are not equal cus we don't know whether the bisector AC is bisecting angle A equally; but how do we establish angle B to be similar to angle A?? did we derive it from the above ratios?
This is a standard figure that begs you to think about similar triangle. Some such figures are given in this post on
Veritas blog:
https://www.veritasprep.com/blog/2014/0 ... -the-gmat/Here, triangle CBA is similar to CDA and both are similar to triangle ABD.
If you want to see how, note this:
Between triangles CBA and ABD,
Angle C is 90 and angle BAD are 90 each. Angle B is common to both. So by AA, both triangles are similar.
Between triangles CDA and ADB,
Angles ACD and BAD are 90 each. Angle D is common to both. So by AA, both triangles are similar.
Between triangles CBA and CAD,
Angle C is 90.
Now say angle BAC is x, then angle CAD is (90-x). Then in triangle CAD, angle D must be x.
Hence angle BAC = angle D
So by AA, both triangles are similar.[/quote]
Oh... understood now.