Question

Fig 5.10 $\square$ABCD is a kite. AB = AD and CB = CD.
Prove that
(i) diagonal AC $\perp$ diagonal BD and
(ii) diagonal AC $\perp$ bisects diagonal BD.

Answer 1

Consider $△$ABC and $△$ADC.
AB = AD (Given)
BC= DC (Given)
CA = CA (Common)

By SSS congruency, $△$ABC $\cong$ $△$ADC
$\angle$DCA = $\angle$BCA (by c.a.c.t)

Now, consider $△$DCP and $△$BCP
DC = CB (Given)
$\angle$DCP = $\angle$BCP (Proved above)
CP = CP (Common)

By SAS congruency, $△$DCP $\cong$ $△$ BCP
$\angle$CPD = $\angle$CPB (by c.a.c.t)
DP = BP (by c.s.c.t)

(i) $\angle$CPD = $\angle$CPB (Proved above)

$\angle$CPD + $\angle$CPB = 180$°$ (Linear pair)
or,
2$\angle$CPD = 180$°$
or,
$\angle$CPD = 90$°$
Hence, diagonal AC is perpendicular to diagonal BD

(ii) DP = BP (Proved above)
So, P is the midpoint of DB.
Hence, AC bisects BD.