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

Answer 2

Answer:

AAA Rule, hence is proved

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