Question

in δabc, l, m, n are points on side ab, bc, ac respectively. perpendicular drawn at l, m, n from δpqr. prove that δabc ~ δpqr.

Answer 1

Answer:

Here triangle ΔABC

      Let ∠A = a,  ∠B =b  and ∠C = c

Now come to quadrilateral ALPN, where

      ∠ ANP =∠ ALP = 90°     (Given)

We know that sum of all interior angle of quadrilateral is 360°

      ∠LAN + ∠LPN +∠ ANP +∠ ALP = 360°

       On putting respective value in above equation

       a + ∠LPN +90° +90° = 360°

       a + ∠LPN = 360° -90° -90°

       a + ∠LPN = 180°

       So

       ∠LPN = 180° - a = ∠LPQ    ...1)

Apply linear pair angle on line LR

      ∠LPQ + ∠RPQ = 180°

      From equation 1)

       180° - a + ∠RPQ = 180°

       So

       ∠RPQ = a          ...2)

Now come to quadrilateral CNQM, where

      ∠ CNQ =∠ QMC = 90°     (Given)

We know that sum of all interior angle of quadrilateral is 360°

      ∠MCN + ∠NQM +∠ CNQ +∠ QMC = 360°

       On putting respective value in above equation

       c + ∠NQM +90° +90° = 360°

       c + ∠NQM= 360° -90° -90°

        c + ∠NQM= 180°

       So

       ∠NQM= 180° - c = ∠NQR    ...3)

Apply linear pair angle on line PN

      ∠NQR + ∠PQR = 180°

      From equation 3)

       180° - c + ∠PQR = 180°

       So

       ∠PQR = c          ...4)

Now come to quadrilateral BMRL, where

      ∠ BLR =∠ RMB = 90°     (Given)

We know that sum of all interior angle of quadrilateral is 360°

      ∠MBL +∠ MRL +∠ BLR +∠ RMB  = 360°

       On putting respective value in above equation

       b + ∠ MRL +90° +90° = 360°

       b + ∠ MRL= 360° -90° -90°

       b + ∠ MRL= 180°

       So

       ∠ MRL= 180° - b = ∠ MRP   ...5)

Apply linear pair angle on line QM

      ∠ MRP + ∠QRP = 180°

      From equation 5)

       180° - b + ∠QRP = 180°

       So

       ∠QRP = b          ...6)

Now from equation 2), equation 4) and equation 6) it is clear that in triangle ΔABC and ΔPRQ

      ∠A = ∠P =a

      ∠B = ∠ R = b

      ∠ C = ∠ Q = c

      So we can say that

             (AAA rules)   Proved