Altitudes AD and BE of acute triangle ABC intersect at point H. If angle AHB = 123 degrees and angle BAH = 26 degrees, then what is angle HCA in degrees?
Asymptote code below
[asy]
size(150); defaultpen(linewidth(0.8));
pair B = (0,0), C = (3,0), A = (1.1,2), P = foot(A,B,C), Q = foot(B,A,C),H = intersectionpoint(B--Q,A--P);
draw(A--B--C--cycle);
draw(A--P^^B--Q);
label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$D$",P,S); label("$E$",Q,NE); label("$H$",H,NW);
[/asy]
Please, there's a message board for a reason
To find angle HCA, we can use the fact that the sum of the angles in a triangle is 180 degrees.
Since AD is an altitude, angle BAH is a right angle. Therefore, angle BAC + angle CAH = 90 degrees.
Similarly, since BE is an altitude, angle AHB is a right angle. Therefore, angle ABC + angle CBH = 90 degrees.
We are given that angle BAH = 26 degrees and angle AHB = 123 degrees.
To find angle HCA, we can subtract the sum of angle BAH and angle ABC from 90 degrees:
angle HCA = 90 degrees - (angle BAH + angle ABC)
angle HCA = 90 degrees - (26 degrees + angle ABC)
Now, we need to find angle ABC. In triangle ABC, angle BCA can be found by subtracting the sum of angle BAC and angle ACB from 180 degrees:
angle BCA = 180 degrees - (angle BAC + angle ACB)
Since angle ACB + angle CAH = 90 degrees (as discussed earlier), we can substitute this value in:
angle BCA = 180 degrees - (angle BAC + 90 degrees - angle BCA)
Simplifying,
angle BCA = 180 degrees - angle BAC - 90 degrees + angle BCA
angle BCA = 90 degrees - angle BAC + angle BCA
Now, we can substitute the given values:
angle BCA = 90 degrees - 26 degrees + angle BCA
angle BCA = 64 degrees + angle BCA
Subtracting angle BCA from both sides,
0 = 64 degrees
This means that angle BCA is 0 degrees, which is not possible in a triangle.
Therefore, there is no solution for the given conditions.
AOPS PROE.
31
The altitudes of a triangle intersect at the orthocenter of the triangle. So, when we extend $\overline{CH}$ to point $F$ on $\overline{AB}$, we have $\overline{CF}\perp\overline{AB}$ as shown below.
[asy]
size(150); defaultpen(linewidth(0.8));
pair B = (0,0), C = (3,0), A = (1.1,2), P = foot(A,B,C), Q = foot(B,A,C),H = intersectionpoint(B--Q,A--P);
draw(A--B--C--cycle);
draw(A--P^^B--Q);
label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$D$",P,S); label("$E$",Q,NE); label("$H$",H+(0,-0.23),SW);
pair F = foot(C,A,B);
draw(C--F);
label("$F$",F,NW);
draw(rightanglemark(C,P,H,3.5));
draw(rightanglemark(H,F,A,3.5));
draw(rightanglemark(H,Q,C,3.5));
[/asy]
From $\triangle HBA$, we have $\angle HBA = 180^\circ - 123^\circ - 26^\circ = 31^\circ$, and from right triangle $BEA$, we have
\[\angle HBA = \angle ABE = 90^\circ - \angle BAE=90^\circ - \angle BAC.\]
But from right triangle $CFA$, we have
\[\angle HCA = \angle FCA = 90^\circ - \angle FAC = 90^\circ - \angle BAC,\]which matches our expression for $\angle HBA$. So, we have $\angle HCA = \angle HBA = \boxed{31^\circ}$.
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