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Can you please explain "Property of right triangle: median from right angle is half of the hypotenuse, hence CD=AB/2=10/2=5" in more detail.
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Excellent, and thank you for explaining
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chetan2u
can the pt mentioned abt property of right angle triangle be proven

Imagine the right triangle inscribed in circle. We know that if the right triangle is inscribed in circle, hypotenuse must be diameter, hence half of the hypotenuse is radius. The line segment from the third vertex to the center is on the on the one hand radius of the circle=half of the hypotenuse and on the other hand as it's connecting the vertex with the midpoint of the hypotenuse it's median too.

Attachment:
Math_Tri_inscribed.png

Hope it's clear.

Thank you for the explanation. The inscribed triangle example helped a lot. Regarding STMT 1, my thought process was that the triangle could be 45-45-90 or 30-60-90, which in my mind, could affect the length of CD. However, from your example or more so the visual, I see that the angles don't matter...that the median is determined by the hypotenuse. Thanks for the lesson learned; basic yet important.
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Brilliant question.
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I think I got this wrong. Let's see after drawing the figure we have a triangle with D written on AB side where AD and CD both have a length of 5 so we actually have two right triangles BDC and ADC.

In the second statement if B is 45 then BCD must also be 45 and hence we can calculate CD

Where am I going wrong here?
Thanks
Cheers
J
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I think I got this wrong. Let's see after drawing the figure we have a triangle with D written on AB side where AD and BD both have a length of 5 so we actually have two right triangles BDC and ADC.

In the second statement if B is 45 then BCD must also be 45 and hence we can calculate CD

Where am I going wrong here?
Thanks
Cheers
J

Notice that CD is the median, not the altitude, so we don't know whether triangles BDC and ADC are right angled.

Does this make sense?
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It doesn't make sense - could you please elaborate on why (2) is NS?

Bunuel
jlgdr
I think I got this wrong. Let's see after drawing the figure we have a triangle with D written on AB side where AD and BD both have a length of 5 so we actually have two right triangles BDC and ADC.

In the second statement if B is 45 then BCD must also be 45 and hence we can calculate CD

Where am I going wrong here?
Thanks
Cheers
J

Notice that CD is the median, not the altitude, so we don't know whether triangles BDC and ADC are right angled.

Does this make sense?
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It doesn't make sense - could you please elaborate on why (2) is NS?

Bunuel
jlgdr
I think I got this wrong. Let's see after drawing the figure we have a triangle with D written on AB side where AD and BD both have a length of 5 so we actually have two right triangles BDC and ADC.

In the second statement if B is 45 then BCD must also be 45 and hence we can calculate CD

Where am I going wrong here?
Thanks
Cheers
J

Notice that CD is the median, not the altitude, so we don't know whether triangles BDC and ADC are right angled.

Does this make sense?

To put it simply: we know the length of a base (10) and one angle at the base (45 degrees). There are infinitely many triangles with this properties, so there are indefinitely many lengths of the median to the base possible.
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Just to be sure, I tried doing this with similar triangles and that's not possible because we don't have at least two side lengths, correct?
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