Question

show that in a right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides

Answer 1

Given: A right triangle, △ABC right angled at B.To prove: AC2=AB2+BC2Construction: Draw AD⊥AC.We need to use the following theorem to prove the above result: Theorem1: If a perpendicluar is drawn from the vertex of a right angle of a right triangle to the hypotenuse, then triangles on each side of the perpendicular are similar to each other and to the whole triangle.



Using the theorem1 stated above, we get    △ABD~△ACBSince the corresponding sides of similar triangles are proportional, we have    ADAB=ABAC Cross multiply to get,    AB2=AC×AD       ......(1)Again, using the theorem1 stated above, we get    △BDC~△ABCSince the corresponding sides of similar triangles are proportional, we have    BCAC=DCBC Cross multiply to get,    BC2=AC×DC        ......(2)Add (1)  and (2)  to get,      AB2+BC2=AC×AD+AC×DC                        =AC×(AD+DC)                        =AC×AC                        =AC2Thus it is proved that,    AB2+BC2=AC2     

Answer 2

Given : A right triangle ABC right angled at B.

To prove : AC^2 = AB^2 + BC^2

Construction : Draw BD ⊥ AC

Proof : In ΔADB and Δ ABC

∠ADB = ∠ABC (each 90°)

∠BAD =∠CAB (common)

ΔADB ~ ΔABC (By AA similarity criterion)

AD  ÷ AB  =  AB  ÷ AC (corresponding sides are proportional)

⇒ AB^2 = AD × AC …(i)

Similarly ΔBDC ~ ΔABC

⇒ BC^2 = CD × AC …(ii)

Adding (1) and (2)

AB^2 + BC^2 = AD × AC + CD × AC

⇒ AB^2 + BC^2 = AC × (AD + CD)

⇒ AB^2 + BC^2 = AC^2