triangle klm is formed by connecting the midpoints of the side of triangle hij. the measures of the interior angles of triangle hij are shown. find the measure of ∠ikl. figures not necessarily drawn to scale.

Question

Accepted Answer

# Explanation:
## Step1: Recall Triangle Midsegment Theorem
The midsegment theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. So, $ KL \parallel HJ $, $ KM \parallel IJ $, and $ LM \parallel HI $. This means triangle $ KLM $ is similar to triangle $ HIJ $ by the AA (Angle - Angle) similarity criterion (since corresponding angles are equal because of parallel lines). Also, we can use the property of parallel lines and transversals to find corresponding angles.

## Step2: Find $ \angle HIJ $ first (sum of angles in a triangle)
In triangle $ HIJ $, the sum of interior angles is $ 180^{\circ} $. We know $ \angle H = 39^{\circ} $ and $ \angle J=58^{\circ} $. So, $ \angle HIJ=180^{\circ}-\angle H - \angle J $.
$$
\begin{align*}
\angle HIJ&=180 - 39- 58\
&=180-(39 + 58)\
&=180 - 97\
&=83^{\circ}
\end{align*}
$$

## Step3: Use parallel lines to find $ \angle IKL $
Since $ KM \parallel IJ $ (by midsegment theorem), $ \angle IKL $ and $ \angle HIJ $ are alternate interior angles? Wait, no. Wait, $ K $, $ L $, $ M $ are midpoints. So $ KL \parallel HJ $, $ KM \parallel IJ $, $ LM \parallel HI $. Let's look at the lines: $ KM $ is parallel to $ IJ $, and $ HI $ is a transversal? Wait, maybe better to see that triangle $ KLM $ is similar to triangle $ HIJ $, so corresponding angles are equal. But also, since $ KM \parallel IJ $, $ \angle IKL=\angle HIJ $? Wait, no, $ \angle IKL $ and $ \angle HIJ $: Wait, $ K $ is on $ HI $, $ L $ is on $ IJ $? Wait, the midpoints: $ K $ is midpoint of $ HI $, $ L $ is midpoint of $ IJ $, $ M $ is midpoint of $ HJ $. So $ KL $ is midsegment, so $ KL \parallel HJ $ and $ KL=\frac{1}{2}HJ $. $ KM \parallel IJ $ and $ KM = \frac{1}{2}IJ $, $ LM \parallel HI $ and $ LM=\frac{1}{2}HI $.

Now, $ \angle IKL $: Let's see, $ KM \parallel IJ $, so $ \angle IKL $ and $ \angle HIJ $ are corresponding angles? Wait, $ HI $ is a side, $ K $ is midpoint of $ HI $, $ L $ is midpoint of $ IJ $, $ M $ is midpoint of $ HJ $. So $ KM $ is parallel to $ IJ $, so the transversal $ HI $ creates alternate interior angles? Wait, $ \angle IKL $ and $ \angle HIJ $: Wait, $ \angle HIJ = 83^{\circ} $ (from step 2), and since $ KM \parallel IJ $, $ \angle IKL=\angle HIJ $? Wait, no, $ \angle IKL $ and $ \angle HIJ $: Wait, $ K $ is on $ HI $, $ L $ is on $ IJ $, $ M $ is on $ HJ $. So $ KL $ is midsegment, so $ KL \parallel HJ $, $ KM \parallel IJ $. So $ \angle IKL $ and $ \angle HIJ $: Wait, $ \angle IKL $ is equal to $ \angle HIJ $ because $ KM \parallel IJ $ and $ HI $ is a transversal? Wait, no, $ \angle IKL $ and $ \angle HIJ $: Wait, $ \angle HIJ = 83^{\circ} $, and in the diagram, $ \angle I = 83^{\circ} $. Since $ KM \parallel IJ $, $ \angle IKL=\angle HIJ = 83^{\circ} $? Wait, no, wait, maybe I made a mistake. Wait, let's re - examine.

Wait, the problem is to find $ \angle IKL $. Let's look at the lines: $ KM $ is parallel to $ IJ $, so $ \angle IKL $ and $ \angle HIJ $ are alternate interior angles? Wait, $ HI $ is a side, $ K $ is midpoint of $ HI $, $ M $ is midpoint of $ HJ $, so $ KM \parallel IJ $ (midsegment theorem: segment connecting midpoints of two sides is parallel to the third side). So $ KM \parallel IJ $, and $ HI $ is a transversal. Then $ \angle IKL $ and $ \angle HIJ $ are alternate interior angles? Wait, $ \angle IKL $ is at $ K $ on $ HI $, between $ IK $ and $ KL $, and $ \angle HIJ $ is at $ I $ on $ HI $, between $ HI $ and $ IJ $. So yes, since $ KM \parallel IJ $, $ \angle IKL=\angle HIJ = 83^{\circ} $? Wait, but in the diagram, $ \angle I $ is $ 83^{\circ} $, so that makes sense. Wait, but let's check again.

Wait, triangle $ HIJ $: angles are $ \angle H = 39^{\circ} $, $ \angle J = 58^{\circ} $, so $ \angle I=180 - 39 - 58 = 83^{\circ} $. Now, $ K $ is midpoint of $ HI $, $ L $ is midpoint of $ IJ $, so $ KL $ is midsegment, so $ KL \parallel HJ $, and $ KM $ is midsegment, so $ KM \parallel IJ $. So quadrilateral $ KIJM $ is a parallelogram? Because $ KM \parallel IJ $ and $ KL \parallel HJ $? Wait, no, $ KI $ and $ MJ $ are not necessarily parallel. Wait, $ K $ is midpoint of $ HI $, $ M $ is midpoint of $ HJ $, so $ KM \parallel IJ $ and $ KM=\frac{1}{2}IJ $. $ L $ is midpoint of $ IJ $, so $ IL=\frac{1}{2}IJ $, so $ KM = IL $ and $ KM \parallel IL $, so $ KMLI $ is a parallelogram. Therefore, $ \angle IKL=\angle IML $, but also, in parallelogram $ KMLI $, $ \angle IKL=\angle HIJ $ (since $ KM \parallel IJ $ and $ HI $ is transversal). So $ \angle IKL = 83^{\circ} $.

# Answer:
$ 83^{\circ} $

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QUESTION IMAGE

Question

Explanation:

Step1: Recall Triangle Midsegment Theorem

Step2: Find \( \angle HIJ \) first (sum of angles in a triangle)

Step3: Use parallel lines to find \( \angle IKL \)

Answer: