proof for basic proportionality theorem - Maths - Triangles

Question

proof for basic proportionality theorem

Gursheen kaur. · Accepted Answer

BPT(Thales Theorem) -

Given - In triangle ABC, DE is parallel to BC DE bisects AB at D and AC at E To Prove - AD/BD= AE/CE Construction - Join BE and CD and draw EF is perpendicular to AB and DG is perpendicular to AC Proof - As, EF is perpendicular to AB, EF is the height of tr's ADE and DBE $a r (A D E) = \frac{1}{2} * b * h = \frac{1}{2} * A D * E F a r (D B E) = \frac{1}{2} * B D * E F N o w, \frac{a r (A D E)}{a r (D B E)} = \frac{\frac{1}{2} * A D * E F}{\frac{1}{2} * B D * E F} = \frac{A D}{B D} (i) s i m i l a r l y, \frac{a r (A D E)}{a r (D C E)} = \frac{\frac{1}{2} * A E * D G}{\frac{1}{2} * C E * D G} = \frac{A E}{E C} (i i) b u t t r D B E a n d t r D C E a r e o n t h e s a m e b a s e D E & b / w t h e s a m e p a r a l l e l s t r a i g h t l i n e s B C & D E t h e r e f o r e, a r (D B E) = a r (D C E) (i i i) f r o m (i), (i i) & (i i i), w e g e t, \frac{A D}{D B} = \frac{A E}{E C} h e n c e t h e t h e o r e m$