Example 2: Applying the Angle Bisector Theorems - three geometry problems involving angle bisectors and measurements.

Let’s solve each part step by step.

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Part A: Find LM

We are given a diagram with point M on the angle bisector of ∠JKL. From point M, perpendiculars are drawn to both sides of the angle — one to side KJ (meeting at J) and one to side KL (meeting at L). The length from M to J is labeled as 12.8.

There’s a key theorem here: The Angle Bisector Theorem for distances — if a point lies on the angle bisector, then it is equidistant from the two sides of the angle. That means the perpendicular distance from M to each side is equal.

So, since MJ = 12.8, and ML is also a perpendicular from M to the other side, then:

→ LM = MJ = 12.8

That matches what’s written in red in the image.

✔ So, LM = 12.8

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Part B: Find m∠ABD, given that m∠ABC = 112°

Looking at the diagram: Point D is inside angle ABC. From D, we drop perpendiculars to BA and BC — meeting at A and C respectively. Both perpendicular segments DA and DC are labeled as 74. Also, there are tick marks showing DA ≅ DC.

Again, this uses the same idea: If a point is equidistant from the two sides of an angle, then it lies on the angle bisector.

Since DA = DC and they’re perpendicular to the sides, point D must lie on the angle bisector of ∠ABC.

Therefore, BD bisects ∠ABC.

Given: m∠ABC = 112°

If BD bisects it, then it splits the angle into two equal parts:

→ m∠ABD = 112° ÷ 2 = 56°

✔ So, m∠ABD = 56°

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Part C: Find m∠TSU

In this diagram, point S is connected to U and R. There are perpendiculars from U to ST and SR — UT ⊥ ST and UR ⊥ SR. Also, UT and UR have tick marks meaning they are equal in length.

Again, using the same rule: If a point (here, U) is equidistant from the two sides of an angle (∠TSR), then it lies on the angle bisector.

So, SU bisects ∠TSR.

Now, look at the angles marked:

- ∠UST = (5z + 23)°
- ∠USR = (6z + 14)°

But wait — actually, looking more carefully: The labels are placed near points T and R, but the angles being referred to are likely ∠UST and USR? Or maybe ∠TSU and ∠RSU?

Actually, re-examining: The problem asks for m∠TSU, which is the angle at S between points T, S, U.

And we see that SU is the angle bisector of ∠TSR, so it should split ∠TSR into two equal angles: ∠TSU and ∠RSU.

But in the diagram, the expressions (5z + 23)° and (6z + 14)° are labeled near the base — probably representing the measures of ∠UST and USR? Wait — no, those are angles at S.

Actually, let’s interpret correctly:

The diagram shows:

- At point S, there are two angles formed by line SU: one toward T, one toward R.
- The angle between TS and SU is labeled (5z + 23)° → that’s ∠TSU
- The angle between RS and SU is labeled (6z + 14)° → that’s ∠RSU

But if SU is the angle bisector, then these two angles should be equal!

Wait — that can’t be unless 5z + 23 = 6z + 14.

Let’s set them equal:

5z + 23 = 6z + 14

Subtract 5z from both sides:

23 = z + 14

Subtract 14:

z = 9

Now plug back in to find either angle:

∠TSU = 5z + 23 = 5(9) + 23 = 45 + 23 = 68°

Check the other: 6z + 14 = 6(9) + 14 = 54 + 14 = 68° → same! Good.

So yes, SU bisects the angle, and each half is 68°.

✔ Therefore, m∠TSU = 68°

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Final Answer:
A: 12.8
B: 56°
C: 68°