toAny One)() In the figure, D, E, F are midpointsof sides AB, BC and A

1.	toAny One)() In the figure, D, E, F are midpointsof sides AB, BC and AC respective-ly. P is the foot of the perpendicularfrom A to side BC. Show that pointsD, F, E and P are concyclicP E
Answer» 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

toAny One)() In the figure, D, E, F are midpointsof sides AB, BC and AC respective-ly. P is the foot of the perpendicularfrom A to side BC. Show that pointsD, F, E and P are concyclicP E

Answer»

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