In the quadrilateral ABCD AB = AC = AD = 12.5, BC = 7 and CD = 15. Find BD
Could you show us what you've done so far with the question?
Well, i figured out , by using the cosine theorem ( that we're not supposed to use in this problem ) that the angle ACD=ADC = aprox 53° and the angle ACB=ABC = aprox 74° But that seems uselees since i dont know ABD.
Well first why don't you write down the lengths you know, next to their corresponding sides?
And we will continue from there.
You can use the drawing tool right at the bottom of this reply section.
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Well we can start by proving similar triangles.
ABD and ACD would be ideal but I don't think you can prove the two are similar. We can start off small by proving these two triangles shaded. |dw:1368952878474:dw|
You can add an extra letter in the middle ie. O
Add as in "label" I mean.
I think it's impossible to prove that ABO is similar or congruent to ADO ( O is the intersection of AC and BD) because we have two sides and one angle equals, but BO is clearly not equal to OD, since this is not a paralelogram.
Nope. We can prove that angle ABD is equal to angle ADb becaue they are base angles of the isosceles triangle ABD.
and then we have a common side AO. And AB=AD.
That's SAS.
Wouldn't be SAS only if angles BAO = DAO? And if BO = OD, that means that the diagonals would bisect each other , and that is only possible if ABCD was a parallelogram. And since 7 is not equal to 12.5 , ABCD is not a parallelogram, and is impossible for their diagonals to bisect each other. Isn't that a contradiction
AB=AD in triangle ADB
THat makes ADB an isosceles triangle.
Yes
So therefore AngleADB=AngleABD. That's one reason why ABO is similar to ADO.
Second reason is that they have a common side AO.
And finally AB=AD. The two triangles can be similar or congruent whichever way you want it to be.
Ok, they are similar, so how we use that?
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