Question

In fig 5.9 $\square$ABCD is a square. P and Q are the points such that seg AQ $\cong$ seg DP, Prove that seg AQ $\perp$ seg DP.
[hint: Prove that $∆$DAP $\cong$ $∆$ABQ and $\angle$$\alpha$ + $\angle$$\beta$ = 90$°$ in $∆$ATP]

Answer 1

Consider $△$DAP and $△$ABQ
$\angle$DAP = $\angle$ABQ = 90$°$ ($\because$ABCD is a square)
seg DP $\cong$ seg AQ (Given)
seg DA $\cong$ seg AB ($\because$ABCD is a square)

By RHS congruency criteria, $△$DAP $\cong$ $△$ABQ
$\angle$DPA = $\angle$AQB = $\alpha$ (by c.a.c.t)
$\angle$BAQ = $\angle$ADP = $\beta$ (by c.a.c.t)

Consider $△$ABQ.
$\alpha$ + $\beta$ + $\angle$ABQ = 180$°$ [Angle sum property]
or,
$\alpha$ + $\beta$ = 90$°$ ($\angle$ABQ = 90$°$)

Now, consider $△$APT
$\alpha$ + $\beta$ + $\angle$ATP = 180$°$ [Angle sum property]
or,
$\angle$ATP = 90$°$ ($\alpha$ + $\beta$ = 90$°$)
∴seg AQ$\perp$seg DP