In triangle A B C we have A B = B C , ∠ B = 2 0 ∘ . Point M on A C is such that A M : M C = 1 : 2 , point H is the projection of C to B M . Find ∠ A H B in degrees.
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What is Q? Where is it?
Make sure your notation doesn't do double duty: "Lets assume the mid point of AC is M.". But M is already a point on AC such that A M : M C = 1 : 2 .
I think you swapped notation, where M in the question is your Q, and you defined a new M. This makes reading the solution somewhat confusing. Can you make the corresponding edits?
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Thanks. This is much better now :)
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@Calvin Lin , SIR IS there any website where i can draw physics related diagram to post it along with question at BRILLIANT?
Now i upvoted it. Its understandable now.
@One Top , One thing more,
what if H lies outside the triangle? Then B H D C is not cyclic?
Please tell what to do.
You have given the vertex angle as 20° and the base angles as 80° each. In such a triangle H can only be within the triangle. Would that be right?
Let's say we have a flat isosceles triangle in which B is obtuse. Now it's possible for H to be outside triangle ABC. The entire figure now changes but BHCD is still cyclic and it can be very easily seen that /- ABH = 90° + B/2. Remember this is /-ABH and not /-AHB.
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Lets assume the mid point of AC is D. Then BHDC is cyclic quad. And hence MD.MC = MH.MB Lets assume AM = X, Then MC = 2X, MD = 3X/2. - X = X/2 We THEN have MD.MC = X² = AM^2 or AM² = MD.MC = MH.MB Thus MA is tangential to the circum-circle of Tr. AHB and thus / ABH = / HAM which makes /_AHB=100°.