Question

state and proof pythagoras theorom and midpoint theoram​

Answer 1

Answer:

PYTHAGORAS THEOREM PROOF

( see figure in the image )

Given: A ∆ XYZ in which ∠XYZ = 90°.

To prove: XZ2 = XY2 + YZ2

Construction: Draw YO ⊥ XZ

Proof: In ∆XOY and ∆XYZ, we have,

∠X = ∠X → common

∠XOY = ∠XYZ → each equal to 90°

Therefore, ∆ XOY ~ ∆ XYZ → by AA-similarity

⇒ XO/XY = XY/XZ

⇒ XO × XZ = XY2 ----------------- (i)

In ∆YOZ and ∆XYZ, we have,

∠Z = ∠Z → common

∠YOZ = ∠XYZ → each equal to 90°

Therefore, ∆ YOZ ~ ∆ XYZ → by AA-similarity

⇒ OZ/YZ = YZ/XZ

⇒ OZ × XZ = YZ2 ----------------- (ii)

From (i) and (ii) we get,

XO × XZ + OZ × XZ = (XY2 + YZ2)

⇒ (XO + OZ) × XZ = (XY2 + YZ2)

⇒ XZ × XZ = (XY2 + YZ2)

⇒ XZ 2 = (XY2 + ZY2 )

2 here is SQUARE

MIDPOINT THEOREM PROOF -

The Midpoint Theorem says that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of the third side.

Proof: Through C, draw a line parallel to BA, and extend DE such that it meets this parallel at F, as shown below:

Midpoint Theorem

Compare

Δ AED with ∆ CEF

1. AE = EC (E is the midpoint of AC)

2. ∠DAE = ∠FCE

(alternate interior angles)

3. ∠DEA = ∠FEC

(vertically opposite angles)

By the ASA criterion, the two triangles are congruent. Thus, DE = EF and AD = CF. But AD is also equal to BD, which means that BD = CF (also, BD || CF by our construction). This implies that BCFD is a parallelogram. Thus,

1. DF || BC è DE || BC

2. DE = EF = ½(DF) = ½(BC) èDE = ½(BC)

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Answer 2

HELLO BUDDY ❤

Answer:

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side. ... Parallel sides are shown by using this symbol ||. You also know the line segment is one-half the length of the third side.