If AC bisects angle ABD and CE in an altituede of triangle ABD And CE is an altitude of triangle ACD find the degree measure of each angle.

Question

If AC bisects angle ABD and CE in an altituede of triangle ABD And CE is an altitude of triangle ACD find the degree measure of each angle.

Answer 1

OK, I drew out the description, and notice that... (sorry, unfortunately I cannot draw this out here)... well, we have a problem with the description.

AC cannot bisect ∠ABD. When we speak of an angle∠ABD, we are saying that "B" (in the middle) is the vertex of an angle subtended by 2 other points, A and D, that will be bisected. The only segment that can bisect ABD is going to have a "B" in it. We could say that BC or BE bisects ∠ABC, but... in this description, we are pretty sure about AC bisecting something. We don't (yet) know about C, but we do know A. Therefore...

I suspect what is meant is: AC bisects angle ∠BAD, not ∠ABD (although, note that ΔABD = ΔBAD)... this gives us point C on the segment BD, which lies opposite of vertex A.

Next, we speak of "an altitude" CE, which wants to be an altitude, simultaneously of larger ΔABD and of the smaller ΔACD. We have established that C must be on BD, so for altitude CE, we start at C, and we must go somewhere to find E... we must go to either of the 2 other sides of ΔABD, which are either AB or AD. Since we are interested in an altitude of ΔACD, we must choose E on side AD, but where on AD ?

The problem is now that, with E on AD, CE cannot be an altitude, simultaneously of the smaller ΔACD, and of the larger triangle ΔABD... unless... unless, CE = CA, and therefore E = A (or CD = CE, and E = D, which produces the same result), so we explore this.

CA bisects ∠BAD, and simultaneously becomes an altitude of larger ΔABD, as well as an altitude (a leg in this case) of smaller right triangle ΔACD.

In fact, the altitude CA at C defines 2 right angles ∠BCA and ∠DCA, a common side CA (the altitude), and 2 equal angles (from angle bisection) at ∠BAC and ∠DAC. Therefore ΔACD ≅ ΔACB (congruent) by ASA Postulate. This means that BC = CD become congruent legs of 2 "inner" right triangles, with CA = CE being the shared Leg of both "inner" right triangles, also AB = AD become congruent hypotenuses of the 2 "inner" triangles, each sharing a right angle vertex at C, and so we also have ∠B = ∠D by congruence. But if ∠B = ∠D, then ΔBAD is Isosceles.

In other words, what we have so far, is ΔBAD is Isosceles in "apex angle"∠A, with AB = AD, "base angles" ∡B = ∡D, with altitude at C centered on side BD opposite vertex ∠A. This is all we know so far, from the information given.

[Special Cases] It is tempting to consider what would happen if ∡A = 90º, i.e. ΔBAD is Right Isosceles triangle. Then each congruent triangle ΔACD ≅ ΔACB becomes 45-45-90 triangles ==> ∡A = 90º, ∡B = ∡C = 45º, and that would be an answer, if it were the only possibility. Another example would be ∡A = 60º, ∡B = ∡C = 30º, etc...

But, I see no evidence that would require anything this specific. They are possibilities, not necessities, given the premise (as re-interpreted)... unless I missed some important detail.

Well, it also may be that my re-interpretation of the errors in the description did not produce the originally intended problem to be solved. Difficult to know for sure...