Let ABC be a triangle such that angle ACB is acute. Suppose that D is an interior point of the triangle ABC such that ∡ADB=∡ACB+pi/2 and AC cdot BD=AD cdot BC. (a) Find (AB cdot CD)/(AC cdot BD). (b) Show that the tangents at C to the circumcicle of the triangle ACD and the circumcircle of the triangle BCD are perpendicular.

Rishi Hale 2022-07-21 Answered
Let ABC be a triangle such that A C B is acute. Suppose that D is an interior point of the triangle ABC such that A D B = A C B + π 2 and A C B D = A D B C ..
a) Find A B C D A C B D ..
(b) Show that the tangents at C to the circumcicle of the triangle ACD and the circumcircle of the triangle BCD are perpendicular.
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Answers (2)

Urijah Hahn
Answered 2022-07-22 Author has 13 answers
Step 1
Draw the perpendicular to CB and then choose E on it, s.t. C B = C E, as in the picture below. Now obviously A C E = A D B and also from the condition:
A C C E = A C B C = A D D B .

Step 2
Therefore A C E A D B. So in particular we have C A E = D A B. Also from the similarity of the triangles we have that A C A D = A E A B ..
This gives us that A C D A B E. So using that BCE is a right isosceles triangle we have from A C D A B E that C D A B = E B A D = 2 B C A D = 2 A C B D.
Hence the ratio is 2 .

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Ethen Blackwell
Answered 2022-07-23 Author has 3 answers
Step 1
Let's invert about D eith arbitrary radius r>0. For any point X in the plane let X∗ be the image of X under the inversion. Then, we will rewrite all conditions in terms of A∗, B∗, C∗ and D.
Firstly, A C B = A C D + B C D = D A C + D B C and A D B = A D B , so we have A D B = A D B .
Hence, A C B = π 2 .
Step 2
Secondly, recall that for any points M and N (other than D) we have M N = R 2 D M D N M N   and   D M = R 2 D M , and M N = R 2 D M D N M N   and   D M = R 2 D M ,, so the second equality can be rewritten as R 2 D A D C A C R 2 D B = R 2 D B D C B C R 2 D A ,, or A C = B C ..
Thus, the triangle A B C is isosceles and right-angled with A C B = π 2

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