Given ΔQXY with RS || XY Prove XR/RQ = YS/SQ statements: 1. RS || XY 2. ∠1=∠3, ∠2=∠4 3. ΔQXY ~ΔQRS 4. XQ/RQ = YQ/SQ 5. XQ=XR+RQ, YQ=YS+SQ 6. XR+RQ/RQ = YS=SQ/SQ 7. XR/RQ = YS/SQ

It can be proved by:

The two triangles are similar and the ratio of corresponding sides are equal.

Step-by-step explanation:

We are given a ΔQXY which has a line RS inside it such that RS || XY.

To prove:

First of all, let us have a look at the ΔQXY and ΔQRS as per the given question figure:

1. ∠1=∠3 (Because sides XY || RS and ∠1, ∠3 are corresponding angles)

2. ∠2=∠4 (Because sides XY || RS and ∠2, ∠4 are corresponding angles)

3. ∠Q is common to both the triangles.

So, all the three angles are equal to each other, hence the two triangles are similar:

If two triangles are similar, then ratio of their corresponding sides is also equal.

Subtracting 1 from both the sides:

Now, it is clearly observable that:

XQ - RQ = XR and

YQ - SQ = YS

Putting the values in above equation:

Hence proved.

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there are two angles that are alternate interior angle with angle 9

the angle 15 and angle 7 are alternate interior angle with angle 9.

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