Question

D,E,F are midpoints of sides AB,BC and AC respectively. P is foot of the perpendicular from A to side BC. Show that point D,F,E and P are concyclic

Answer 1

Answer:

Step-by-step explanation:

Given : In ΔABC, D,E and F are the mid points of sides AB, BC, CA respectively. AP ⊥ BC.

To prove : E, F, D and P are concyclic.

Proof :

In ΔABC, D and F are mid points of AB and CA respectively.

∴ DF || BC (Mid point theorem)

Similarly, EF || AB and ED || CA

In quadrilateral BEFD,

BE || DF and EF || BD (DF || BC and EF || AB)

∴ Quadrilateral BEFD is a parallelogram.

Similarly, quadrilateral ADEF is a parallelogram.

∴ ∠A = ∠DEF (Opposite sides of parallelogram are equal)

ED || AC and EC is the transversal,

∴ ∠BED = ∠C (Corresponding angles)

∠DEF = ∠DED + ∠DEF = ∠A + ∠C ...(1)

DF || BC and BD is the transversal,

∴ ∠ADO = ∠B (Corresponding angles) ...(2)

In ΔABD, D is the mid point of AB and OD || BP.

∴ O is the mid point of AP (Converse of mid point theorem)

⇒ OA = OP

In ΔAOD and ΔDOP,

OA = OP (Proved)

∠AOD = ∠DOP (90°) {∠DOP = ∠OPE (Alternate angles) & ∠AOD = ∠DOP = 90° (linear pair)}

OD = OD (Common)

∴ ΔAOD congruence ΔDOP (SAS congruence criterion)

⇒ ∠ADO = ∠PDO (CECT)

⇒ ∠PDO = ∠B (Using (2))

In quadrilateral PDFE,

∠PDO + ∠PEF = ∠B + ( ∠A + ∠C) = ∠A + ∠B + ∠C (Using (1))

⇒ ∠PDO + ∠PEF = 180° ( ∠A + ∠B + ∠C = 180°)

Hence, quadrilateral PDFE is a cyclic quadrilateral.

Thus, the points E, F, D and P are concyclic.

Answer 2

Answer: Proved

Step-by-step explanation:

Given : In Δ ABC , D,E,F are the mid _ point of the sides BC , CA, AND AB  respectively , AP ⊥ BC

TO prove : D,F,E and P are con cyclic.

Proof : In right angled triangle  APB, D is the mid point of AB

                        DB =DP

                 ∠ 2 = ∠1 ....(1)

( angle opposite to the equal sides are equal)

Since D and F are the mid point of AB and AC , then

                                         DF║ BC

                                        DF ║ BE

Since EF ║DB , then quadrilateral BEFD is a parallelogram

                                                 ∠1 = ∠3....(2)

 From equation 1 & 2

                                          ∠2 =∠3

                But              ∠2 + ∠4 = 180 °         ( linear pair Axiom )

∴                                   ∠3 + ∠4 = 180°              (∵ ∠2 =∠3 )

hence , Quadrilateral EFDP is cyclic quadrilateral

  So, points D,E,F and P are con - cyclic