Question

show that in a triangle, the square of the hypotenuse is equal to the sum of the square of the other side

Answer 1

Prove that triangles ABC and CBE are similar.
It follows from the AA postulate that triangle ABC is similar to triangle CBE, since angle B is congruent to angle B and angle C is congruent to angle E. Thus, since internal ratios are equal s/a=a/c.
Multiplying both sides by ac we get
sc=a^2.

Now show that triangles ABC and ACE are similar.
As before, it follows from the AA postulate that these two triangles are similar. Angle A is congruent to angle A and angle C is congruent to angle E. Thus, r/b=b/c. Multiplying both sides by bc we get
rc=b^2.

Now when we add the two results we get
sc + rc = a^2 + b^2.
c(s+r) = a^2 + b^2
c^2 = a^2 + b^2,
concluding the proof of the Pythagorean Theorem.

Answer 2

Answer:

Given :

A right triangle ABC right angled at B.

To prove :

AC² = AB² + BC²

Construction :

Draw BD ⊥ AC

Proof :

In Δ ADB and Δ ABC

∠ A = ∠ A    [ Common angle ]

∠ ADB = ∠ ABC   [ Both are 90° ]

∴  Δ  ADB  Similar to Δ ABC   [ By AA similarity ]

So , AD / AB = AB / AC   [ Sides are proportional ]

= > AB² = AD . AC  ... ( i )

Now in Δ BDC and Δ ABC

∠ C = ∠ C    [ Common angle ]

∠ BDC = ∠ ABC   [ Both are 90° ]

∴  Δ  BDC Similar to Δ ABC   [ By AA similarity ]

So , CD / BC = BC / AC

= > BC² = CD . AC   ... ( ii )

Now adding both equation :

AB² + BC² = CD . AC +  AD . AC

AB² + BC² = AC ( CD + AD )

AB² + BC² = AC² .

AC² = AB² + BC² .

Hence proved .