Question

In fig. 3.23, $\angle$BAC = 40$°$, seg PB $\cong$ seg AB $\cong$ seg AC $\cong$ seg CQ then find $\angle$ PAQ and find all pairs of congruent triangles. Justify your answer.

Answer 1

Given : seg PB $\cong$ seg AB $\cong$ seg AC $\cong$ seg CQ
$\angle$BAC = 40$°$
As seg AB $\cong$ seg AC,
$\angle$ACB = $\angle$ABC … (isosceles triangle theorem)
$\Rightarrow$180$°$ $-$ $\angle$ACQ = 180$°$ $-$ $\angle$ABP ( as they form linear pairs of angles)
$\Rightarrow$ $\angle$ACQ = $\angle$ABP … (1)
Now, in $∆$ABC,
$\angle$BAC = 40$°$
$\angle$ABC + $\angle$ACB + $\angle$BAC = 180$°$ … (by angle sum property)
$\Rightarrow$ 2$\angle$ABC = 180$°$ $-$ $\angle$BAC ....(∵ $\angle$ACB = $\angle$ABC )
$\Rightarrow$ 2$\angle$ABC = 180$°$ $-$ 40$°$
$\Rightarrow$ $\angle$ABC = 70$°$ = $\angle$ACB
$\Rightarrow$$\angle$ABP = $\angle$ACQ = 180$°$ $-$70$°$ = 110$°$ (linear pair)


In $∆$APB,
seg AB $\cong$ seg PB
$\Rightarrow$$\angle$BAP = $\angle$BPA (isosceles triangle theorem) ........ (2)
Similarly in $∆$ACQ,
$\angle$CAQ = $\angle$CQA (isosceles triangle theorem) ...... (3)
Now again in $∆$APB,
$\angle$ABP + $\angle$BAP +$\angle$BPA =180$°$ ….(by angle sum property)
$\Rightarrow$ 2$\angle$BPA =180$°$ $-$ 110$°$ ....{from (2)}
$\Rightarrow$$\angle$BPA = 35$°$ = $\angle$BA
Similarly, $\angle$CQA = $\angle$CAQ = 35$°$
In $∆$APB and $∆$ACQ,
PB = CQ (given)
$\angle$ABP = $\angle$ACQ {from(1)}
AB = AC (given)
So, $∆$ABP $\cong$ $∆$ACQ (SAS congruency test)


seg PB $\cong$seg CQ
$\Rightarrow$ seg PB + seg BC $\cong$seg CQ + seg BC
$\Rightarrow$ seg PC $\cong$ seg BQ … (4)

Now, in $∆$ACP and $∆$ABQ,
AC = AB
$\angle$ACP = $\angle$ABQ
PC = BQ … (from(4))
So, $∆$ACP $\cong$ $∆$ABQ … (SAS congruency test)