Prove that \(\Delta A P B \sim \Delta C P D\) [byAA similarity criterion]
PD = 5 cm, CD = 2 cm

100°

Given \(\triangle NSQ\cong \triangle MTR\) and \(\angle1=\angle2\)
To prove \(\triangle PTS\sim \triangle PRQ\)
Proof Since, \(\triangle NSQ\cong \triangle MTR\)
\(\therefore\) SQ = TR  ... (i)
Also, \(\angle1=\angle2\)
\(\Rightarrow\) PT = PS .... (ii)
[since, sides opposite to equal angles are also equal]
From Eqs.(i) and (ii), \(\frac{PS}{SQ}=\frac{PT}{TR}\)
\(\Rightarrow ST\parallel QR\)
[by converse of basic proportionality theorem]
\(\therefore \angle 1=\angle PQR\)   and \( \angle 2=\angle PRQ\)
[\(\therefore\)  atternate exterior angle]
In \(\triangle PTS\) and  \(\triangle PRQ\),
\(\angle P=\angle P\)    [common angle]
\(\angle 1=\angle PQR\)  [proved above]
and \(\angle 2=\angle PRQ\)
\(\therefore \triangle PTS\sim \triangle PRQ\)
[by AAA similarity criterion]

A trapezium ABCD in which \(PQ\parallel AB\), draw a line AC, which intersects PQ at O.
Join AC, which intersects PQ at O.
Given, \(AB\parallel DC\) and \(PQ\parallel AB\)
Then, \(AB\parallel PQ\parallel DC\)

In \(\triangle ADC\), \(PO\parallel DC\)
By basic proportionality theorem,
\(\frac{AP}{PD}=\frac{AO}{OC}\)  ... (i)
In \(\triangle CAB\), \(OQ\parallel AB\)
By basic proportionality theorem,
\(\frac{AO}{OC}=\frac{BQ}{QC}\) ... (ii)
On comparing Eqs. (i) and (ii), we get
\(\frac{AP}{PD}=\frac{BQ}{QC} \Rightarrow \frac{AP}{18}=\frac{35}{15}\)
[\(\because\)  PD = 18 cm, BQ = 35 cm and QC = 15 cm]
\(\Rightarrow AP=\frac { 18\times 35 }{ 15 } \) = 42 cm
\(\therefore\)  AD = AP + PO
= 42 + 18 = 60 cm

Apply Pythagoras theorem, and use the result p - q = 1