Question

state and prove bpt therom and Pythagoras therom​

Answer 1

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Answer 2

BPT THEORM :

o prove this theorem first we will join BE and CD. Then draw a line EL perpendicular to AB and line DM perpendicular to AC. Now we will find the ratio of area of Δ

ADE to Δ

DBE and ratio of area of Δ

ADE to Δ

ECD. Comparing the ratios we will get the final answer.

Complete step-by-step answer:

Now, ΔDBE

and ΔECD

being on the same base DE and between the same parallels DE and BC, we have,ar(ΔDBE)=ar(ΔECD)

then we say that the basic proportionality theorem is proved.

Basic proportionality theorem:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

Given:

A ΔABC

in which DE∥BC

and DE intersects AB and AC at D and E respectively.

To prove that:

ADDB=AEEC

Construction:

Join BE and CD.

Draw EL⊥AB

and DM⊥AC

Proof:

We have the

ar(ΔADE)=12×AD×EL

ar(ΔDBE)=12×DB×EL

Therefore the ratio of these two is ar(ΔADE)ar(ΔDBE)=ADDB

. . . . . . . . . . . . . . (1)

Similarly,

ar(ΔADE)=ar(ΔADE)=12×AE×DM

ar(ΔECD)=12×EC×DM

Therefore the ratio of these two is ar(ΔADE)ar(ΔECD)=AEEC

. . . . . . . . . . . .. . . (2)

Now, ΔDBE

and ΔECD

being on the same base DE and between the same parallels DE and BC, we have,

ar(ΔDBE)=ar(ΔECD)

. . . . . . . . . . . (3)

From equations 1, 2, 3 we can conclude that

ADDB=AEEC

Hence we can say that the basic proportionality theorem is proved.

Note: The formula for area of the triangle is given by 12×b×h

where b, h are base and height respectively. If two triangles are on the same base and between the same parallels then the area of those two triangles are equal.

PYTHAGORAS THEROM : Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

The sides of the right-angled triangle are called base, perpendicular and hypotenuse.

 

According to Pythagoras theorem,

(AC)2=(AB)2 + (BC)2(AC)2=(AB)2 + (BC)2

Proof:

Given, a triangle ABC in which ∠ABC is 900∠ABC is 900.

Construction: Draw a perpendicular BD on AC i.e., BD ⊥⊥ AC.

 

In ΔABD and ΔABC ΔABD and ΔABC  we have,

∠BAD = ∠BAC ∠BAD = ∠BAC i.e. ∠A∠A is common in both triangles.

∠ABC = ∠ADB = 900∠ABC = ∠ADB = 900

Therefore ΔABC∼ΔABD ΔABC∼ΔABD ( By AA similarity i.e. angle-angle similarity)

So,⇒ADAB=ABAC⇒AB2 = AD×AC ...(1)⇒ADAB=ABAC⇒AB2 = AD×AC ...(1)

In ΔBDC and ΔABC ΔBDC and ΔABC  we have,

∠BCD = ∠BCA ∠BCD = ∠BCA i.e. ∠C∠C is common in both triangles.

∠ABC = ∠ADC = 900∠ABC = ∠ADC = 900

Therefore ΔABC∼ΔBDC ΔABC∼ΔBDC ( By AA similarity i.e. angle-angle similarity)

So,⇒DCBC=BCAC⇒BC2 = AC×DC ...(2)⇒DCBC=BCAC⇒BC2 = AC×DC ...(2)

Adding equation (1) and (2) , we get

⇒AB2 + BC2 = AD×AC + AC× DC⇒AB2 + BC2 = AC(AD + DC)⇒AB2 + BC2 = AC(AC)⇒AB2 + BC2 = AC2⇒AB2 + BC2 = AD×AC + AC× DC⇒AB2 + BC2 = AC(AD + DC)⇒AB2 + BC2 = AC(AC)⇒AB2 + BC2 = AC2

Hence, proved.

Note: In a right angled triangle , hypotenuse is the longest side of the triangle and is opposite to the right angle i.e. 900900. By drawing a perpendicular from point B and dividing the triangle ABC into 2 parts and using angle-angle similarity to prove the Pythagoras theorem.

HOPE IT HELPS YOU.