Question

The exterior angle at vertex C of triangle ABC is

112°. If the bisector of side AB intersects side AC at point Q and line segment BQ
is perpendicular to AC, find the measure of angle ABC​

Answer 1

Explanation:

We know that the exterior angle at vertex C is equal to the sum of the interior angles at vertices A and B that are not adjacent to C. Thus, we have:

∠ACB = ∠A + ∠B

Since ∠AQC is the angle bisector of ∠BAC, we have:

∠AQD = ∠CQB

Since BQ is perpendicular to AC, we have:

∠BQC = 90° - ∠QCB

Now, let x be the measure of angle ABC. Then we have:

∠A = 180° - ∠ACB - x

∠B = 180° - ∠ACB - (90° - ∠QCB)

Substituting the given values and simplifying, we get:

∠A = 68° - x

∠B = 22° + ∠QCB

Now, we use the fact that the angles in a triangle add up to 180°:

∠A + ∠B + ∠C = 180°

Substituting the above expressions and simplifying, we get:

68° - x + 22° + ∠QCB + 112° = 180°

Simplifying further, we get:

∠QCB + x = 22°

But we also know that ∠QCB + x is equal to ∠ABC. Thus, we have:

∠ABC = 22°

Therefore, the measure of angle ABC is 22 degrees.