Question

In the given figure if D,E and F are mid points of sides AB,AC and BC respectively and AG is perpendicular to BC. then prove that DEFG is a cyclic quadrilateral

Answer 1

Answer:

Proof

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