In the adjacent figure in which AD is perpendicular to BC and AB = p, AC = q, BD = r, DC = s then show that p^2-q^2=r^2-s^2
Answers
Answer:
Given Two triangles ABC and DBC lie on the same side of the base BC. Points P,Q and R are points on BC,AC and CD respectively such that PR||BD and PQ||AB.
To prove QR||AD
Proof In △ABC, we have
PQ∣∣AB
∴
PB
CP
=
QA
CQ
........(i) [By Basic proportionality Theorem]
In △BCD, we have
PR∣∣BD
∴
PB
CP
=
RD
CR
........(ii) [By Thale's Theorem]
From (i) and (ii), we have
QA
CQ
=
RD
CR
Thus, in △ACD, Q and R are points on AC and CD respectively such that
QA
CQ
=
RD
CR
⇒ QR∣∣AD [By the converse of Basic Proportionality Theorem]
Answer:
Given Two triangles ABC and DBC lie on the same side of the base BC. Points P,Q and R are points on BC,AC and CD respectively such that PR||BD and PQ||AB.
To prove QR||AD
Proof In △ABC, we have
PQ∣∣AB
∴
PB
CP
=
QA
CQ
........(i) [By Basic proportionality Theorem]
In △BCD, we have
PR∣∣BD
∴
PB
CP
=
RD
CR
........(ii) [By Thale's Theorem]
From (i) and (ii), we have
QA
CQ
=
RD
CR
Thus, in △ACD, Q and R are points on AC and CD respectively such that
QA
CQ
=
RD
CR
⇒ QR∣∣AD [By the converse of Basic Proportionality Theorem]
Step-by-step explanation: