Proving ( AD frac{BC}{2} ) in a Right-Angled Triangle

In this article, we will explore the geometric properties of a right-angled triangle and prove why, in a right-angled triangle ( ABC ), if ( D ) is the midpoint of the hypotenuse ( BC ), then ( AD frac{BC}{2} ). We will present several proofs, including the use of similar triangles, Thales' theorem, and congruent triangles.

Proof Using Similar Triangles

Consider a right-angled triangle ( ABC ) with right angle at ( A ). Let ( D ) be the midpoint of the hypotenuse ( BC ).

Let the lengths of the sides of the triangle be: ( AB a ) ( AC b ) ( BC c ) Since ( triangle ABC ) is a right-angled triangle, by the Pythagorean theorem: [ a^2 b^2 c^2 ] Given that ( D ) is the midpoint of ( BC ), we have: [ BD DC frac{c}{2} ] We need to prove that ( AD frac{c}{2} ). Using the similar triangles ( triangle ABD ) and ( triangle ADC ): [ frac{AB}{AD} frac{BD}{DC} ] Simplifying, we get: [ frac{a}{AD} frac{frac{c}{2}}{frac{c}{2}} Rightarrow frac{a}{AD} 1 Rightarrow AD a ] Rewriting ( a ) in terms of ( sqrt{a^2 b^2} ) (which is ( c )): [ AD a frac{c}{sqrt{a^2 b^2}} frac{c}{c} frac{BC}{2} ]

Proof Using Thales' Theorem

Thales' theorem states that if a line segment is drawn through the midpoints of the two sides of a triangle, it is parallel to the third side and half as long.

Note that Thales' theorem directly applies to this scenario. Given ( triangle ABC ) with right angle at ( A ), and ( D ) as the midpoint of ( BC ), we can state: ( AD ) is the radius of the circle with ( BC ) as its diameter as per Thales' theorem. ( AD BD DC frac{BC}{2} ). Thus, we prove that ( AD frac{BC}{2} ).

Proof Using Congruent Triangles

Consider the following construction:

Through ( D ), draw perpendiculars ( DE ) and ( DF ) to ( AC ) and ( AB ) respectively, meeting ( AB ) and ( AC ) at ( E ) and ( F ). To prove: ( DE ) and ( DF ) are the perpendicular bisectors of ( AB ) and ( AC ). Proof: By construction: [ FD perp AB ] and [ DE perp AC ]. Triangles ( triangle CFD ) and ( triangle BED ) are congruent because: ( angle CFD angle BED 90^circ ) ( angle FDC angle EBD ) (corresponding angles) ( CF DE ) and ( DF EB ) Since ( FAED ) is a parallelogram: [ DE FA ] and ( AE FD ). Therefore, ( AF FC ) and ( AE EB ). Thus, ( E ) and ( F ) are the midpoints of ( AB ) and ( AC ). Hence, ( DE ) and ( DF ) are perpendicular bisectors of ( AB ) and ( AC ). This proves that ( AD BD DC frac{BC}{2} ).

Conclusion

The proofs presented demonstrate why, in a right-angled triangle, if ( D ) is the midpoint of the hypotenuse, then ( AD ) is half the length of the hypotenuse. This property holds true for all right-angled triangles and is useful in various geometric and trigonometric applications.

Keywords

right-angled triangle, midpoint, hypotenuse, Thales' theorem