Question

Derive Appolonius Theorm .


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Answer 1

Answer:

Appllonius theorem:

Statement:

In ΔABC, if M is the midpoint of side BC, then

AB² + AC² = 2AM² + 2BM².

Step-by-step-explanation:

NOTE: Kindly refer to the attachment first.

Given:

In ΔABC, M is the midpoint of BC.

∴ BM = CM

To prove:

AB² + AC² = 2AM² + 2BM²

Construction:

Draw seg AD ⊥ side BC such that B-D-C.

Proof:

In first case, AM is not perpendicular to side BC, then out of ∠ AMB and ∠ AMC, one is acute angle and other is obtuse angle.

Let's consider ∠ AMB as acute angle and ∠ AMC as obtuse angle.

Δ ABM is an acute angled triangle.

∴ By application of Pythagoras' theorem,

AB² = AM² + BM² - 2BM.DM .... ( 1 )

Δ AMC is an obtuse angled triangle.

∴ By application of Pythagoras' theorem,

AC² = AM² + MC² + 2MC.DM .... ( 2 )

But, BM = CM ... [ Given ] ( 3 )

Substituting ( 3 ) in ( 2 ), we get,

AC² = AM² + BM² + 2BM.DM .... ( 4 )

Adding ( 1 ) and ( 4 ), we get,

AB² + AC² =

AM² + BM² - 2BM.DM + AM² + BM² + 2BM.DM

∴ $\boxed{AB^{2} + AC^{2} = 2AM^{2} + 2BM^{2}}$

Now, in second case, seg AM ⊥ seg BC.

∴ In Δ AMB, ∠ AMB = 90°.

∴ AB² = AM² + CM² ... [ Pythagoras theorem ] ( 5 )

In Δ AMC, ∠ AMC = 90°

∴ AC² = AM² + CM² ... [ Pythagoras theorem ] ( 6 )

But, BM = CM .... [ Given ] ( 7 )

∴ AC² = AM² + BM² .... [ From ( 6 ) & ( 7 ) ]

( 8 )

Adding ( 5 ) and ( 8 ), we get,

AB² + AC² = AM² + BM² + AM² + BM²

∴ $\boxed{AB^{2} + AC^{2} = 2AM^{2} + 2BM^{2}}$

Additional Information:

1. Appllonius Theorem:

It was given by a mathematician Appllonius.

This theorem shows the relation between the sides and medians of the triangles.

2. Pythagoras Theorem:

It was given by a mathematician Pythagoras.

This theorem shows the relation between the longest side ( hypotenuse ) and the remaining two sides in a right-angled triangle.