Question

circle touches the side EF of ∆DEF at P and

touches sides DE and DF at Q and R resp. when

produced. Show that DQ= ½(perimeter of ∆DEF)​

Answer 1

Answer:

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.⇒AQ=AR,BQ=BP,CP=CR.

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.⇒AQ=AR,BQ=BP,CP=CR.Perimeter of △ABC=AB+BC+CA

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.⇒AQ=AR,BQ=BP,CP=CR.Perimeter of △ABC=AB+BC+CA=AB+(BP+PC)+(AR–CR)

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.⇒AQ=AR,BQ=BP,CP=CR.Perimeter of △ABC=AB+BC+CA=AB+(BP+PC)+(AR–CR)=(AB+BQ)+(PC)+(AQ–PC) since [AQ=AR,BQ=BP,CP=CR]

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.⇒AQ=AR,BQ=BP,CP=CR.Perimeter of △ABC=AB+BC+CA=AB+(BP+PC)+(AR–CR)=(AB+BQ)+(PC)+(AQ–PC) since [AQ=AR,BQ=BP,CP=CR]=AQ+AQ

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.⇒AQ=AR,BQ=BP,CP=CR.Perimeter of △ABC=AB+BC+CA=AB+(BP+PC)+(AR–CR)=(AB+BQ)+(PC)+(AQ–PC) since [AQ=AR,BQ=BP,CP=CR]=AQ+AQ=2AQ

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.⇒AQ=AR,BQ=BP,CP=CR.Perimeter of △ABC=AB+BC+CA=AB+(BP+PC)+(AR–CR)=(AB+BQ)+(PC)+(AQ–PC) since [AQ=AR,BQ=BP,CP=CR]=AQ+AQ=2AQ⇒AQ=

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.⇒AQ=AR,BQ=BP,CP=CR.Perimeter of △ABC=AB+BC+CA=AB+(BP+PC)+(AR–CR)=(AB+BQ)+(PC)+(AQ–PC) since [AQ=AR,BQ=BP,CP=CR]=AQ+AQ=2AQ⇒AQ= 2

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.⇒AQ=AR,BQ=BP,CP=CR.Perimeter of △ABC=AB+BC+CA=AB+(BP+PC)+(AR–CR)=(AB+BQ)+(PC)+(AQ–PC) since [AQ=AR,BQ=BP,CP=CR]=AQ+AQ=2AQ⇒AQ= 21

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.⇒AQ=AR,BQ=BP,CP=CR.Perimeter of △ABC=AB+BC+CA=AB+(BP+PC)+(AR–CR)=(AB+BQ)+(PC)+(AQ–PC) since [AQ=AR,BQ=BP,CP=CR]=AQ+AQ=2AQ⇒AQ= 21

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.⇒AQ=AR,BQ=BP,CP=CR.Perimeter of △ABC=AB+BC+CA=AB+(BP+PC)+(AR–CR)=(AB+BQ)+(PC)+(AQ–PC) since [AQ=AR,BQ=BP,CP=CR]=AQ+AQ=2AQ⇒AQ= 21 (Perimeter of △ABC)

Given:A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.⇒AQ=AR,BQ=BP,CP=CR.Perimeter of △ABC=AB+BC+CA=AB+(BP+PC)+(AR–CR)=(AB+BQ)+(PC)+(AQ–PC) since [AQ=AR,BQ=BP,CP=CR]=AQ+AQ=2AQ⇒AQ= 21 (Perimeter of △ABC)∴AQ is the half of the perimeter of △ABC