Question

in triangle abc l m n are points on side a b BC AC respectively perpendicular drawn at element from triangle pqr prove that triangle ABC similar to triangle prq​

Answer 1

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Answer 2

Step-by-step explanation:

• 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

        $\mathbf{\Delta ABC\sim \Delta PRQ}$       (AAA rules)   Proved