In the figure, AB ||CD To find the value of x and y complete the following activity: ABCD & PQ is the transversal ZX- Reason Zy= Reason
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In ΔABD and ΔPQD
∠D= ∠D (common)
∠BAD=∠QPD (corresponding angles)
So, ΔABD∼ΔPQD (by AA similarity criterion)
⇒
PQ
AB
=
QD
BD
(Ratio of corresponding sides of two similar triangles)
⇒
z
x
=
QD
BQ+QD
⇒
z
x
=
QD
BQ
+1
z
x
−1=
QD
BQ
⇒
z
x−z
=
QD
BQ
---(i)
Similarly in ΔCBD and ΔPBQ
∠B=∠B (common)
∠BCD=∠BPQ (corresponding angles)
So,ΔCBD∼ΔPBQ (by AA similarity criterion)
⇒
PQ
CD
=
BQ
BD
(Ratio of corresponding sides of two similar triangles)
⇒
z
y
=
BQ
BQ+QD
⇒
z
y
=1+
BQ
QD
⇒
z
y−z
=
BQ
QD
⇒
y−z
z
=
QD
BQ
--(ii)
from (i) and (ii)
z
x−z
=
y−z
z
⇒(x−z)(y−z)=z
2
⇒xy−yz−xz+z
2
=z
2
xy=xz+yz
⇒
x
1
+
y
1
=
z
1
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